Kinematics 2D
2024-09-11
www.njctl.org
Table of Contents:
Kinematics 2D
Click on a topic to go to that section
· Adding Vectors in Two Dimensions
· Basic Vector Operations
· Galilean Transformations
· Projectile Motion Type 1
· Projectile Motion Type 2
· Projectile Motion Type 3
· Calculating Landing Velocity
Adding Vectors in
Two Dimensions
https://njctl.org/video/?v=m5B9pWdvDV0
Return to
Table of
Contents
Adding Vectors
Previously, we learned about how
displacement, velocity and
acceleration are vector quantities.
North
Vectors can be added together.
Let's walk through adding 2
displacement vectors together.
West
East
South
Adding Vectors
Adding two displacement vectors:
North
1. Draw the first vector,
beginning at the origin, with its
tail at the origin.
East
West
South
Adding Vectors
Adding two displacement vectors:
North
1. Draw the first vector,
beginning at the origin, with its
tail at the origin.
2. Draw the second vector with
its tail at the tip of the first
vector.
West
East
-y
South
Adding Vectors
Adding two displacement vectors:
North
1. Draw the first vector,
beginning at the origin, with its
tail at the origin.
2. Draw the second vector with
its tail at the tip of the first
vector.
3. Draw the resultant (the
answer) from the tail of the first
vector to the tip of the last.
West
East
South
Adding Vectors
The direction of each vector matters.
+y
In this first case, the vector sum of:
5 units to the East plus
2 units to the West is
+x
-x
3 units to the East.
-y
Adding Vectors
For instance, if the second vector
had been 2 units to the EAST
(not west), we get a different
answer.
In this second case, the vector
sum of:
+y
+x
-x
5 units to the East plus
2 units to the EAST is
7 units to the East .
-y
Adding Vectors
Vectors that point in the same direction add together and create an
even longer vector:
+
=
While vectors that point in opposite directions subtract from each
other:
+
=
Remember, the magnitude of a vector refers to its length. Here's
an example with numbers:
7
+
-5
=
2
Adding Vectors in 2-D
But how about if the vectors are
along different axes?
North
What is the vector sum of:
5 units East plus
2 units North
West
East
South
Adding Vectors
1. Draw the first vector,
beginning at the origin, with
its tail at the origin.
North
East
West
South
Adding Vectors
1. Draw the first vector,
beginning at the origin, with
its tail at the origin.
North
2. Draw the second vector with
its tail at the tip of the first
vector.
West
East
South
Adding Vectors
1. Draw the first vector,
beginning at the origin, with
its tail at the origin.
North
2. Draw the second vector with
its tail at the tip of the first
vector.
West
3. Draw the resultant (the
answer) from the tail of the
first vector to the tip of the
last.
East
South
The resultant vector ends up being the hypotenuse of a right triangle.
Adding Vectors
Calculating the resultant
vector's magnitude and
direction requires the use of
right triangle mathematics
(trigonometry).
North
c
West
b
a
East
South
We know the length of both SIDES of the triangle (a and b), but we
need to know the length of the HYPOTENUSE (c).
Magnitude of a Resultant
The magnitude of the resultant
is equal to the length of the
vector.
North
We get the magnitude of the
resultant from the
Pythagorean theorem:
c=?
b = 2 units
West
c =a +b
c2 = 5 2 + 2 2
c2 = 25 + 4
c2 = 29
2
2
a=5
units
2
c = √ (29) = 5.4 units
South
East
Adding Vectors
In physics, we say the
direction of a vector is equal to
the angle θ between a chosen
axis and the resultant.
North
With 2-D vectors, we can't just say
the answer is positive or negative
to signify east or west (or north or
West
south) like we could in one
dimension.
We will primarily use the x-axis to
measure θ, though in certain
situations, we will measure θ with
respect to the y-axis.
θ
East
South
Adding Vectors
To find the value of the angle θ,
we need to use what we already
know:
North
the length of the two sides
opposite and adjacent to the
angle.
(remember SOH CAH TOA)
θ
West
East
South
Adding Vectors
tan(θ) = opp / adj
North
tan(θ) = (2 units) / (5 units)
tan(θ) = 2/5
tan(θ) = 0.40
θ
West
East
To find the value of θ, we must
take the inverse tangent:
θ = tan -1 (0.40) = 22 0
South
Adding Vectors
North
5.4
θ
220
West
East
South
The Resultant is 5.4 units in the direction of 22 0 North of East
magnitude
direction
1 What is the magnitude of the Resultant of two vectors
A and B, if A = 8.0 units north and B = 4.5 units east?
A 9.2 units
C 12.5 units
D 14.0 units
E I need help
https://njctl.org/video/?v=5RPwCWLmHH8
Answer
B 10.4 units
1 What is the magnitude of the Resultant of two vectors
A and B, if A = 8.0 units north and B = 4.5 units east?
A 9.2 units
B 10.4 units
Answer
C 12.5 units
A
D 14.0 units
E I need help
[This object is a pull
tab]
https://njctl.org/video/?v=5RPwCWLmHH8
2 What is the magnitude of the Resultant of two vectors
A and B, if A = 24.0 units east and B = 15.0 units
south?
A 24.1 units
C 39.0 units
D 44.9 units
E I need help
https://njctl.org/video/?v=6paEhH7xxMg
Answer
B 28.3 units
2 What is the magnitude of the Resultant of two vectors
A and B, if A = 24.0 units east and B = 15.0 units
south?
A 24.1 units
Answer
B 28.3 units
C 39.0 units
B
D 44.9 units
E I need help
[This object is a pull
tab]
https://njctl.org/video/?v=6paEhH7xxMg
3 What is the direction of the resultant of the two vectors
A and B if: A = 8.0 units north and B = 4.5 units east if
East is 0o and North is 90o?
A 34.8o
C 53.6o
D 60.6o
E I need help
https://njctl.org/video/?v=x-cw4jAna-U
Answer
B 41.4o
3 What is the direction of the resultant of the two vectors
A and B if: A = 8.0 units north and B = 4.5 units east if
East is 0o and North is 90o?
A 34.8o
B 41.4o
Answer
C 53.6o
D 60.6o
E I need help
https://njctl.org/video/?v=x-cw4jAna-U
D
[This object is a pull
tab]
4 What is the direction (from East) of the Resultant of the
two vectors A and B if: A = 24.0 units east and B = 15.0
units south?
A 13o South of East
C 40o South of East
D 49o South of East
E I need help
https://njctl.org/video/?v=kVgTBIlCAL0
Answer
B 32o South of East
4 What is the direction (from East) of the Resultant of the
two vectors A and B if: A = 24.0 units east and B = 15.0
units south?
A 13o South of East
Answer
B 32o South of East
C 40o South of East
B
D 49o South of East
E I need help
https://njctl.org/video/?v=kVgTBIlCAL0
[This object is a pull
tab]
5 Find the magnitude of the resultant of two vectors
A and B if:
Magnitude = ?
A 150 units
B 396 units
C 472 units
D 650 units
E I need help
https://njctl.org/video/?v=rl1oAqZ9ZDE
Answer
A = 400 units north
B = 250 units east
5 Find the magnitude of the resultant of two vectors
A and B if:
Answer
A = 400 units north
B = 250 units east
Magnitude = ?
C
A 150 units
B 396 units
C 472 units
D 650 units
E I need help
https://njctl.org/video/?v=rl1oAqZ9ZDE
[This object is a pull tab]
6 Find the direction of the resultant of
two vectors A and B if:
Direction = ?
A 27o East of North
B 32o East of North
C 41o East of North
D 58o East of North
E I need help
https://njctl.org/video/?v=24HRrcMPL8Q
Answer
A = 400 units north
B = 250 units east
6 Find the direction of the resultant of
two vectors A and B if:
A = 400 units north
B = 250 units east
Answer
Direction = ?
A 27o East of North
B
B 32o East of North
C 41o East of North
[This object is a pull tab]
D 58o East of North
E I need help
https://njctl.org/video/?v=24HRrcMPL8Q
A
B
C
D
E
300 m
400 m
500 m
700 m
I need help
https://njctl.org/video/?v=LNJqxYmCbVs
Answer
7 A student walks a distance of 300 m East, then
walks 400 m North. What is the magnitude of the
net displacement?
7 A student walks a distance of 300 m East, then
walks 400 m North. What is the magnitude of the
net displacement?
300 m
400 m
500 m
700 m
I need help
Answer
A
B
C
D
E
C
[This object is a pull
tab]
https://njctl.org/video/?v=LNJqxYmCbVs
A
B
C
D
E
1090 m
700 m
1700 m
1300 m
I need help
https://njctl.org/video/?v=jAqP_yrdEqo
Answer
8 A student walks a distanceof 500 m East, then
walks 1200 m North. What is the magnitude of
the net displacement?
8 A student walks a distanceof 500 m East, then
walks 1200 m North. What is the magnitude of
the net displacement?
1090 m
700 m
1700 m
1300 m
I need help
Answer
A
B
C
D
E
D
[This object is a pull tab]
https://njctl.org/video/?v=jAqP_yrdEqo
A
B
C
D
is 2.0 m.
could be as small as 2.0 m, or as large as 12 m.
is 12 m.
is larger than 12 m.
E
I need help
https://njctl.org/video/?v=H7Ma1NY2hpg
Answer
9 Two displacement vectors have magnitudes of 5.0 m
and 7.0 m, respectively. W hen these two vectors are
added, the magnitude of the sum:
9 Two displacement vectors have magnitudes of 5.0 m
and 7.0 m, respectively. W hen these two vectors are
added, the magnitude of the sum:
is 2.0 m.
could be as small as 2.0 m, or as large as 12 m.
is 12 m.
is larger than 12 m.
E
I need help
Answer
A
B
C
D
B
[This object is a pull
tab]
https://njctl.org/video/?v=H7Ma1NY2hpg
A 0°
B 45°
C 90°
D 180°
E I need help
https://njctl.org/video/?v=ZNg5OgbQDn8
Answer
10 The resultant of two vectorsis the largest when the
angle between them is
10 The resultant of two vectorsis the largest when the
angle between them is
Answer
A 0°
B 45°
C 90°
D 180°
E I need help
A
[This object is a pull
tab]
https://njctl.org/video/?v=ZNg5OgbQDn8
A
B
C
D
E
0°
45°
90°
180°
I need help
https://njctl.org/video/?v=7WiUKeXXOHc
Answer
11 The resultant of two vectors is the smallest
when the angle between them is:
11 The resultant of two vectors is the smallest
when the angle between them is:
0°
45°
90°
180°
I need help
Answer
A
B
C
D
E
D
[This object is a pull
tab]
https://njctl.org/video/?v=7WiUKeXXOHc
Basic Vector Operations
https://njctl.org/video/?v=XAoHOZnQQvs
Return to
Table of
Contents
Adding Vectors
Adding Vectors in the opposite order gives the same resultant.
V1 + V2 = V2 + V1
North
North
V1
V2
V2
West
V1
South
East
West
East
South
Adding Vectors
Even if the vectors are not at right angles, they can be added
graphically by using the "tail to tip" method.
The resultant is drawn from the tail of the first vector to the tip
of the last vector.
V1
+
V2
+
V3
V1
=
V2
VR
V3
Adding Vectors
...and the order in which you add them does not matter.
V1
V3
+
V1
+
V2
=
V3
VR
V2
Subtracting Vectors
In order to subtract a vector, we add the negative of that
vector. The negative of a vector is defined as that vector in
the opposite direction.
V1
-
=
V2
V1
-V2
=
VR
V1
+
-V2
Multiplication of Vectors by Scalars
A vector V can be multiplied by a scalar c. The result is a
vector cV which has the same direction as V. However, if c
is negative, it changes the direction of the vector.
V
2V
-½V
Adding Vectors by Components
Any vector can be described as the sum of two other vectors
called components. These components are chosen
perpendicular to each other and can be found using
trigonometric functions.
y
Vy
V
θ
Vx
x
Adding Vectors by Components
In order to remember the right triangle properties and to
better identify the functions, it is often convenient to show
these components in different arrangements (notice v y below).
y
V
Vy
θ
Vx
x
Adding Vectors by Components
Using the tip-to-tail method, we can sketch the resultant of
any two vectors.
y
V
V2
V1
x
But we cannot find the exact magnitude of the resultant, v,
since
v1 and v 2 in the sketch are not accurate nor precise enough.
Adding Vectors by Components
We now know how to break v 1 and v 2 into components...
y
V
2y
V2
V1
V2x
V1y
V1x
x
Adding Vectors by Components
And since the x and y components are one dimensional,
they can be added as such.
y
V2y
Vy = v1y + v2y
V1y
V1x
V2x
Vx = v 1x + v 2x
x
Adding Vectors by Components
1.
v2
V1
Draw a diagram and add the
vectors graphically.
Adding Vectors by Components
y
v2
V1
x
1.
Draw a diagram and add the
vectors graphically.
2.
Choose x and y axes.
Adding Vectors by Components
y
V2y
v2
V1
V2x
V1y
V1x
x
1.
Draw a diagram and add the
vectors graphically.
2.
Choose x and y axes.
3.
Resolve each vector into x
and y components.
Adding Vectors by Components
y
V2y
v2
V1
V2x
1.
Draw a diagram and add the
vectors graphically.
2.
Choose x and y axes.
3.
Resolve each vector into x
and y components.
4.
Calculate each component.
V1y
V1x
x
v1x = v1 cos(θ1) v2x = v2 cos(θ2)
v1y = v1 sin(θ1) v2y = v2 sin(θ2)
Adding Vectors by Components
y
Vx
V2y
v2
Vy
V1
V2x
1.
Draw a diagram and add the
vectors graphically.
2.
Choose x and y axes.
3.
Resolve each vector into x
and y components.
4.
Calculate each component.
5.
Add the components in each
direction.
V1y
V1x
x
Adding Vectors by Components
y
Vx
V
Vy
V2y
v2
V1
V2x
1.
Draw a diagram and add the
vectors graphically.
2.
Choose x and y axes.
3.
Resolve each vector into x
and y components.
4.
Calculate each component.
5.
Add the components in each
direction.
6.
Find the length and direction of
the resultant vector.
V1y
V1x
x
Example 1: Adding Vectors by Components
Graphically determine the resultant of the following three vector
displacements:
1. 24m, 30º north of east
2. 28m, 37º east of north
3. 20m, 50º west of south
Example 1: Adding Vectors by Components
Graphically determine the resultant of the following three vector
displacements:
1. 24m, 30º north of east
2. 28m, 37º east of north
3. 20m, 50º west of south
28m
First, draw the vectors.
37O
24m
30O
20m
50O
Example 1: Adding Vectors by Components
Graphically determine the resultant of the following three vector
displacements:
1. 24m, 30º north of east
2. 28m, 37º east of north
3. 20m, 50º west of south
28m
Next, find the x and y
components of each vector.
37O
24m
30O
20m
50O
Example 1: Adding Vectors by Components
Graphically determine the resultant of the following three vector
displacements:
1. 24m, 30º north of east
2. 28m, 37º east of north
3. 20m, 50º west of south
28m
Next, find the x and y
components of each vector.
37O
24m
30O
20m
50O
Example 1: Adding Vectors by Components
Graphically determine the resultant of the following three vector
displacements:
1. 24m, 30º north of east
2. 28m, 37º east of north
3. 20m, 50º west of south
28m
Next, find the x and y
components of each vector.
37O
24m
30O
20m
The vectors are both negative
because south and west are
negative directions.
50O
Example 1: Adding Vectors by Components
Graphically determine the resultant of the following three vector
displacements:
1. 24m, 30º north of east
2. 28m, 37º east of north
3. 20m, 50º west of south
x (m)
y (m)
d1
20.8
12.0
d2
16.9
22.4
d3
-15.3
-12.9
Ʃ
22.4
21.5
To find the magnitude of the
resultant, use the Pythagorean
theorem.
To find the direction of the
resultant, use inverse tangent.
A
B
C
D
E
Translate it parallel to itself
Rotate it
Multiply it by a constant factor
Add a constant vector to it
I need help
https://njctl.org/video/?v=lZwiy4eUJfY
Answer
12 Which of the following operations will not change a
vector?
12 Which of the following operations will not change a
vector?
Translate it parallel to itself
Rotate it
Multiply it by a constant factor
Add a constant vector to it
I need help
Answer
A
B
C
D
E
A
[This object is a pull
tab]
https://njctl.org/video/?v=lZwiy4eUJfY
A
B
C
D
E
12 m/s
15 m/s
20 m/s
25 m/s
I need help
https://njctl.org/video/?v=UeJpcY4YEQ8
Answer
13 If a ball is thrown with a velocity of 25 m/s at an
angle of 37° above the horizontal, what is the vertical
component of the velocity?
13 If a ball is thrown with a velocity of 25 m/s at an
angle of 37° above the horizontal, what is the vertical
component of the velocity?
12 m/s
15 m/s
20 m/s
25 m/s
I need help
Answer
A
B
C
D
E
B
[This object is a pull
tab]
https://njctl.org/video/?v=UeJpcY4YEQ8
A
B
C
D
E
12 m/s
15 m/s
20 m/s
25 m/s
I need help
https://njctl.org/video/?v=VW68wDtVumE
Answer
14 If a ball is thrown with avelocity of 25 m/s at an angle
of 37° above the horizontal, what is the horizontal
component of thevelocity?
14 If a ball is thrown with avelocity of 25 m/s at an angle
of 37° above the horizontal, what is the horizontal
component of thevelocity?
12 m/s
15 m/s
20 m/s
25 m/s
I need help
Answer
A
B
C
D
E
C
[This object is a pull tab]
https://njctl.org/video/?v=VW68wDtVumE
A
B
C
D
E
19°
45°
60°
71°
I need help
https://njctl.org/video/?v=lSkU8cit_GU
Answer
15 If you walk 6.0 km in a straight line in a direction
north of east and you end up 2.0 km north and
several kilometers east. How many degrees north
of east have you walked?
15 If you walk 6.0 km in a straight line in a direction
north of east and you end up 2.0 km north and
several kilometers east. How many degrees north
of east have you walked?
19°
45°
60°
71°
I need help
Answer
A
B
C
D
E
https://njctl.org/video/?v=lSkU8cit_GU
A
[This object is a pull
tab]
16 Algebraically determine the magnitude of the
resultant of the following three vector displacements:
1. 15 m, 30º north of east
2. 20 m, 37º north of east
A 60.1 m
B 72.4 m
C 85.3 m
D 92.0 m
E I need help
https://njctl.org/video/?v=fWcRlt77zdA
Answer
3. 25 m, 45o north of east
16 Algebraically determine the magnitude of the
resultant of the following three vector displacements:
1. 15 m, 30º north of east
2. 20 m, 37º north of east
Answer
3. 25 m, 45o north of east
A 60.1 m
B 72.4 m
C 85.3 m
D 92.0 m
E I need help
https://njctl.org/video/?v=fWcRlt77zdA
A
[This object is a pull
tab]
17 Algebraically determine the direction of the resultant of
the following three vector displacements:
1. 15 m, 30º north of east
2. 20 m, 37º north of east
A 29.4o
B 38.6o
C 42.9o
D 50.3o
E I need help
https://njctl.org/video/?v=YLuQQo42YFg
Answer
3. 25 m, 45o north of east
17 Algebraically determine the direction of the resultant of
the following three vector displacements:
1. 15 m, 30º north of east
2. 20 m, 37º north of east
3. 25 m, 45o north of east
Answer
A 29.4o
B 38.6o
C 42.9o
D 50.3o
E I need help
https://njctl.org/video/?v=YLuQQo42YFg
B
[This object is a pull
tab]
A
B
C
A vector cannot have a magnitude of zero if
one of its components is not zero.
The magnitude of a vector can be equal to less
than the magnitude of one of its components.
If the magnitude of vector A is less than the
magnitude of vector B, then the x-component
of A must be less than the x-component of B.
D
The magnitude of a vector can be either
positive or negative.
E
I need help
https://njctl.org/video/?v=zkd868s6pI4
Answer
18 Which of the following is an accurate
statement?
18 Which of the following is an accurate
statement?
B
C
A vector cannot have a magnitude of zero if
one of its components is not zero.
The magnitude of a vector can be equal to less
than the magnitude of one of its components.
If the magnitude of vector A is less than the
A
magnitude of vector B, then the x-component
of A must be less than the x-component of B.
Answer
A
D
The magnitude of a vector can be either
positive or negative.
E
I need help
https://njctl.org/video/?v=zkd868s6pI4
[This object is a pull
tab]
Galilean
Transformations
Return to Table
of Contents
https://njctl.org/video/?v=irArqyRzjb0
Reference Frames
Reference frames and relative motion were introduced in the
Kinematics 1D unit of this course.
These concepts will now be expanded so that calculations
can be made to relate the position, velocity and acceleration
of objects in different reference frames.
They will also be expanded to motion in two dimensions.
Reference Frames
In order to describe where something is, or its motion, you
need to relate it to something else.
That's the purpose of a reference frame - choose a
coordinate system in space and make measurements relative
to the system.
There is no one preferred reference frame - and frequently
there is a need to translate measurements made in one
frame to another.
One reference frame can even be moving relative to the
other.
As long as the frames move with a constant velocity relative
to each other, we can use a Galilean transformation.
Accelerating reference frames are the province of General
Relativity and won't be covered in this course.
Reference Frames
In order to describe where something is, or its motion, you
need to relate it to something else.
That's the purpose of a reference frame - choose a
coordinate system in space and make measurements relative
to the system.
There is no one preferred reference frame - and frequently
there is a need to translate measurements made in one
frame to another.
One reference frame can even be moving relative to the
other.
As long as the frames move with a constant velocity relative
to each other, we can use a Galilean transformation.
Accelerating reference frames are the province of General
Relativity and won't be covered in this course.
Reference Frames
Before the math is done, think about different reference
frames.
If you're sitting in a bus that is going 25 m/s (56 mph), the
person sitting next to you, relative to a reference frame that
has you at the origin, is not changing his position, and his
speed relative to you is 0 m/s.
If someone is standing at a bus stop, relative to a reference
system that has her at the origin, the person in the bus is
moving at 25 m/s and is getting farther away from her.
And if a bus is moving parallel to your bus with the same
velocity, a person in that bus would agree with you. Your seat
mate looks stationary.
Who is correct? Why, you all are!
Galilean Transformations
Choose a two dimensional reference frame based on Cartesian
coordinates; we'll only be solving problems in one dimension. The
s (stationary) frame is not moving.
Then, choose the m (moving) system and put it in motion with a
constant velocity, u, (velocity of the m system with respect to the s
system), in the x direction.
The coordinates of an object in the s frame are (x, y) and the
coordinates in the m system are (x', y').
y
s
y'
m
x
u
x'
Galilean Transformations
Lets add a bus which is attached to the m frame.
There's an observer at the origin of the s system.
The bus (and m frame) is moving at a speed, u, with respect to the
stationary observer in the s system. The bus is at rest with respect
to the m frame.
u is the relative velocity of the two frames (observers in both
frames agree on this velocity, but each will express it in the
opposite direction).
y
s
y'
m
x
u
x'
This is a simpler example as
only one of the reference
frames is moving.
Galilean Transformations
Add a person inside the bus, moving at a velocity, vm, relative to
the bus. The person is also moving at a velocity, vs, with respect
to the stationary observer.
The Galilean transformations enable the calculation of the
position, velocity, and acceleration of the person in both frames.
This only works for inertial reference frames - both frames need
to be inertial, that is, they are moving at constant velocities.
y
s
y'
m
x
u
vm
x'
All those motion
parameters are vectors.
Since we're only working
in one dimension, vectors
to the left are negative
and vectors to the right
are positive.
Galilean Transformations
The Galilean position transformation is derived from the second
kinematics equation with a = 0.
x = x' + ut
x is the position of the bus rider in the stationary frame.
x' is the position of the bus rider in the moving frame.
The velocity transformation is:
vs = v m + u
vs is the velocity of the bus rider in the stationary frame.
vm is the velocity of the bus rider in the moving frame.
The acceleration transformation is:
as = a m
y
s
y'
m
x
u
vm
x'
Example 1: Galilean Relativity
A person is walking with a speed of 1.3 m/s from the front of a bus
to the back of a bus. The bus is moving at a speed of 22 m/s.
What speed does an observer at a street corner observe for the
person in the bus?
y
s
y'
m
u
vm
x
x'
answer on next page
Example 1: Galilean Relativity
A person is walking with a speed of 1.3 m/s from the front of a bus
to the back of a bus. The bus is moving at a speed of 22 m/s.
What speed does an observer at a street corner observe for the
person in the bus?
Given: vm = -1.3 m/s u = 22 m/s
Find vs
Use the Galilean transformation for velocity, vs = vm + u. Use a
negative value for vm, since the person in the bus is moving to the
left.
y
s
y'
m
u
vm
x
x'
Example 2: Galilean Relativity
A person is sitting in a bus moving at a speed of 16 m/s. The
person is sitting at a position of 4.2 m from the back of the bus.
The back of the bus passes an observer sitting on a park bench
at t = 0 s. After 5.0 s, at what position does the observer see the
bus rider?
y
s
y'
m
u
x
x'
answer on next page
Example 2: Galilean Relativity
A person is sitting in a bus moving at a speed of 16 m/s. The
person is sitting at a position of 4.2 m from the back of the bus.
The back of the bus passes an observer sitting on a park bench
at t = 0 s. After 5.0 s, at what position does the observer see the
bus rider?
Given: u = 16 m/s x' = 4.2 m t = 5.0 s Find x.
Use the Galilean transformation for position, x = x' + ut. Use a
positive value for x', since the person in the bus is in front of
the back of the bus.
y
The bus rider, relative to
the bus, is still at position
x' = 4.2 m.
s
y'
m
x
u
x'
Two Dimensional Galilean Transformations
For an object in a reference frame (m) that is moving at a speed
of u in the x direction, the Galilean transformations relative to a
stationary frame (s) are:
x = x' + ut
x is the position of the object in the stationary frame.
x' is the position of the object in the moving frame.
vs = vm + u
as = a m
If an object also has a position in the y direction in the moving
frame, one more transformation is added:
y = y'
Observers in both reference frames measure the same y
position, velocity and acceleration of the object.
Two Dimensional Galilean Transformations
A common problem involves a boat crossing a river.
The water is flowing parallel to the river banks and the boat
captain intends to sail perpendicular to the banks to land at a
spot directly across from the boat's starting point.
The problem will be solved with the information provided in this
unit involving the Galilean transformation and vector addition.
Example 3: Galilean Relativity
Two kids are on a power boat capable of a maximum speed of
4.2 m/s, and wish to cross a river 1200 m to a point directly across
from their starting point. The speed of the water in the river is
2.8 m/s.
a. How much time is required for the crossing (assume the engine
is operating at maximum speed)?
b. What angle, θ relative to the line connecting the two points on
the river banks is required?
answer on next pages
Example 3: Galilean Relativity
a. How much time is required for the crossing (assume the engine
is operating at maximum speed)?
Given: u = 2.8 m/s y = y' = 1200 m vmax = 4.2 m/s Find t
River bank
B
y
s
y'
m
The xy frame is fixed to the river
bank. The water flow is
represented by the moving x'y'
frame at speed u.
u
The boat starts at point A and
needs to reach point B.
River
A
x
River bank
A person sitting on the river bank
at point A will see the boat moving
straight across the river.
x'
continued on next pages
Example 3: Galilean Relativity
a. How much time is required for the crossing (assume the engine
is operating at maximum speed)?
Given: u = 2.8 m/s y = y' = 1200 m vmax = 4.2 m/s Find t
If the boat points directly at point
B, the water will push it to the
right, and the boat will never get to
point B.
River bank
B
y
s
y'
m
u
The boat captain should aim the
bow (front) of his boat to the left.
River
A
x
River bank
x'
In this way, even though the boat
is aimed as indicated on the
picture, it will actually move in a
straight line from A to B as
observed by the person on the
river bank at point A.
continued on next pages
Example 3: Galilean Relativity
a. How much time is required for the crossing (assume the engine
is operating at maximum speed)?
Given: u = 2.8 m/s y = y' = 1200 m vmax = 4.2 m/s Find t
Create a vector diagram
with three different
velocities:
· vmax - the maximum
velocity of the boat
(stationary rest frame).
· u - the velocity of the
water current (moving
rest frame)
· var - the velocity the boat
makes across the river as
observed by the person
on the shore (stationary
reference frame).
River bank
u
B
var
y
s
y'
vmax
m u
River
θ
A
x
River bank
x'
Example 3: Galilean Relativity
a. How much time is required for the crossing (assume the engine
is operating at maximum speed)?
Given: u = 2.8 m/s y = y' = 1200 m vmax = 4.2 m/s Find t
Vector addition shows that the
vectors, vmax and u, add
together to find the vector var.
The three velocities make a
right trangle, so use the
Pythagorean theorem to find var. vmax
River bank
u
B
var
y
s
y'
m u
River
θ
A
x
River bank
x'
Example 3: Galilean Relativity
a. How much time is required for the crossing (assume the engine
is operating at maximum speed)?
Given: u = 2.8 m/s y = y' = 1200 m vmax = 4.2 m/s Find t
According to the Galilean
River bank
transformation, the distance in
the y direction traveled by the
u B
y
boat in the moving frame is
y'
v
ar
s
m u
equal to the distance traveled in
the stationary frame, Δy' = Δy.
vmax
The time is also the same, t' = t.
River
θ
A
x
River bank
x'
Example 3: Galilean Relativity
b. What angle, θ relative to the line connecting the two points on
the river banks is required?
Given: u = 2.8 m/s y = y' = 1200 m vmax = 4.2 m/s Find t
The three velocity vectors
make a right triangle, so use
trigonometry.
River bank
u
B
var
y
s
y'
vmax
m u
River
θ
A
x
River bank
x'
Example 3: Galilean Relativity
u
B
vmax
var
Bow
θ
Velocity vector triangle
B
A
Stern
A
The ovals represent the boat as it moves across the river.
The observer on the first bank sees the boat move from
point A to B.
The boat points to the left as it work against the current as if it was sliding across the river.
19 Galilean transformations can be used in which of the
following reference frames?
A merry-go-round
C race car driving around a curve
D race car moving with a constant velocity
E I need help
https://njctl.org/video/?v=sYTHl8qKiz8
Answer
B airplane taking off
19 Galilean transformations can be used in which of the
following reference frames?
A merry-go-round
Answer
B airplane taking off
C race car driving aroundDa curve
D race car moving with a constant velocity
E I need help
[This object is a pull tab]
https://njctl.org/video/?v=sYTHl8qKiz8
20 A boat is moving with a constant velocity diagonally
across a flowing river. Which of the following values for
the boat is the same in the moving boat frame and a
reference frame fixed to the river bank?
B position parallel to the river banks
C velocity parallel to the river banks
D position diagonally across the river
E I need help
https://njctl.org/video/?v=FrqYnwizniA
Answer
A velocity perpendicular to the river banks
20 A boat is moving with a constant velocity diagonally
across a flowing river. Which of the following values for
the boat is the same in the moving boat frame and a
reference frame fixed to the river bank?
A velocity perpendicular to the river banks
Answer
B position parallel to the river banks
C velocity parallel to the river banks
A
D position diagonally across the river
E I need help
[This object is a pull tab]
https://njctl.org/video/?v=FrqYnwizniA
A 2.5 m/s
B 5.9 m/s
C 8.4 m/s
D 10.9 m/s
E I need help
https://njctl.org/video/?v=k_4tESgB8Yk
Answer
21 A sailor is jogging at a speed of 2.5 m/s towards the bow
(front) of an aircraft carrier that is moving at a constant
speed of 8.4 m/s. A merchant seaman on a nearby
container ship at anchor observes the sailor jogging at
what speed?
21 A sailor is jogging at a speed of 2.5 m/s towards the bow
(front) of an aircraft carrier that is moving at a constant
speed of 8.4 m/s. A merchant seaman on a nearby
container ship at anchor observes the sailor jogging at
what speed?
Answer
A 2.5 m/s
B 5.9 m/s
D
C 8.4 m/s
D 10.9 m/s
E I need help
https://njctl.org/video/?v=k_4tESgB8Yk
[This object is a pull tab]
Answer
22 A sailor is jogging on the deck of an aircraft carrier that is
moving at a constant speed of 8.4 m/s. The jogger starts
sprinting towards the bow (front) of the ship at a rate of
3.1 m/s2. What is the acceleration of the sailor observed
by a merchant seaman on a nearby container ship
moving at a constant speed of 4 m/s?
A less than
B equal to
C greater than
D cannot calculate; the time of the acceleration is
required
E I need help
https://njctl.org/video/?v=YxzJv9REdBM
Answer
22 A sailor is jogging on the deck of an aircraft carrier that is
moving at a constant speed of 8.4 m/s. The jogger starts
sprinting towards the bow (front) of the ship at a rate of
3.1 m/s2. What is the acceleration of the sailor observed
by a merchant seaman on a nearby container ship
moving at a constant speed of 4 m/s?
A less than
B equal to
C greater than
B
D cannot calculate; the time of the acceleration is
required
E I need help
https://njctl.org/video/?v=YxzJv9REdBM
[This object is a pull tab]
E I need help
https://njctl.org/video/?v=mZAyTdcxj5A
Answer
23 A person in a descending parachute sees the earth
moving toward at him at a constant velocity v. A
stationary observer on the ground watches the
parachutist glide down with a constant speed of -v.
Which of the people are in an inertial reference frame?
A parachutist only
B ground observer only
C neither the parachutist nor the ground observer
D both the parachutist and the ground observer
Answer
23 A person in a descending parachute sees the earth
moving toward at him at a constant velocity v. A
stationary observer on the ground watches the
parachutist glide down with a constant speed of -v.
Which of the people are in an inertial reference frame?
A parachutist only
B ground observer only
C neither the parachutist nor the ground observer
D
D both the parachutist and the ground observer
E I need help
[This object is a pull tab]
https://njctl.org/video/?v=mZAyTdcxj5A
Projectile Motion
Type 1
https://njctl.org/video/?v=tojs0uX8nWk
Return to
Table of
Contents
Projectile Motion
a=g
A projectile is an object
moving in two dimensions
where it only experiences
an acceleration downward
due to gravity.
vx
v
vy
1
Projectile
Motion
Three types of projectile
motion will be analyzed.
Vertical
fall
There is zero acceleration
in the horizontal direction.
vy
Their paths are parabolas.
vy
v
v vx = v
vx
vx
vy
v
vx
vx
vy
2
vy
vy
v
vy
v
3
vx
vx
v
vx = v
vx
vy
v
vx
vx
vy
v
v
Projectile Motion
a=g
Projectile motion is analyzed by
solving the vertical and horizontal
motion separately.
vx
The speed in the x direction is
constant, as there is zero
acceleration.
v
vy
1
Projectile
Motion
Vertical
fall
The speed in the y direction is
changing as it experiences an
acceleration due to gravity.
vy
vy
v
v vx = v
vx
vx
vy
v
vx
vx
vy
2
vy
vy
v
vy
v
3
vx
vx
v
vx = v
vx
vy
v
vx
vx
vy
v
v
Projectile Motion
a=g
If not for gravity, projectiles would
continue in a straight line, and if
aimed parallel to or above the
horizontal, would never hit the
ground.
vx
Gravity determines how far the
projectile travels, so most
problems start off with solving for
the time the projectile is in the air,
using kinematics equations in the
y direction.
v
vy
1
Projectile
Motion
Vertical
fall
vy
vy
v
v vx = v
vx
vx
vy
v
vx
vx
vy
2
vy
vy
v
vy
v
3
vx
vx
v
vx = v
vx
vy
v
vx
vx
vy
v
v
Projectile Motion
All projectile motion problems will be solved by resolving
them into two separate one dimensional problems (x and y).
Kinematics equations only work for one dimension at a time.
When dealing with the x dimension, use:
And when dealing with the y
dimension, use:
Example 2: Projectile Motion Type 1
A mountain lion leaps horizontally from a 7.5 m high rock with a
speed of 4.5 m/s. How far from the base of the rock will the lion land?
v0
answer on next pages
Example 2: Projectile Motion Type 1
A mountain lion leaps horizontally from a 7.5 m high rock with a
speed of 4.5 m/s. How far from the base of the rock will the lion land?
Analyze the givens:
v0
In the x direction, the initial velocity is
4.5 m/s, there is zero acceleration, and
the initial position is 0.
In the y direction, the acceleration
equals -g, the initial velocity is zero,
and the initial position is 7.5 m.
As the lion falls vertically, the vertical
speed increases. However, due to zero
acceleration in the x direction, the lion's
horizontal speed remains constant.
continued on next page
Example 2: Projectile Motion Type 1
A mountain lion leaps horizontally from a 7.5 m high rock with a
speed of 4.5 m/s. How far from the base of the rock will the lion land?
Many times, drawing a sketch, and adding the givens is helpful. It
doesn't have to be as artistic as this one.
y(m)
vy0 = 0.0 m/s
y0=7.5 m
vx0 = 4.5 m/s
x0 = 0
x=?
y=0
continued on next page
x(m)
Example 2: Projectile Motion Type 1
A mountain lion leaps horizontally from a 7.5 m high rock with a
speed of 4.5 m/s. How far from the base of the rock will the lion land?
Translate the givens into math variables:
y(m)
vy0 = 0.0 m/s
y0=7.5 m
vx0 = 4.5 m/s
x0 = 0
x=?
y=0
x(m)
x
y
x0 = 0 m
x=?
vx0 = 4.5 m/s
vx = 4.5 m/s
t=?
ax = 0 m/s2
y0 = 7.5 m
y=0m
vy0 = 0 m/s
vy = ?
t=?
ay = -9.8 m/s2
Time is the only variable that both directions have in common, and
will always be the piece the connects them.
continued on next page
Example 2: Projectile Motion Type 1
A mountain lion leaps horizontally from a 7.5 m high rock with a
speed of 4.5 m/s. How far from the base of the rock will the lion land?
y
y0 = 7.5 m
y=0m
vy0 = 0 m/s
vy = ?
t=?
a = -9.8 m/s2
Since y0, y, v0y and ay are known, the 2nd kinematics
equation in the y direction to find time. Time is the only
unknown in the equation.
0
Solve the equation for t and substitute in givens.
negative signs
cancel
continued on next page
Example 2: Projectile Motion Type 1
A mountain lion leaps horizontally from a 7.5 m high rock with a
speed of 4.5 m/s. How far from the base of the rock will the lion land?
Now that we have the time the lion is in the air, add it to the list of
givens. Even though we're working in two dimensions, the time
that the lion hits the ground is the same for both - there's only
one lion.
y
x
x0 = 0 m
x=?
vx0 = 4.5 m/s
vx = 4.5 m/s
t = 1.24 s
ax = 0 m/s2
y0 = 7.5 m
y=0m
vy0 = 0 m/s
vy = ?
t = 1.24 s
ay = -9.8 m/s2
The only missing variable is the distance traveled, x.
continued on next page
Example 2: Projectile Motion Type 1
A mountain lion leaps horizontally from a 7.5 m high rock with a
speed of 4.5 m/s. How far from the base of the rock will the lion land?
Use the second kinematics in the x dimension to solve for x, the
final x position.
x
x0 = 0 m
x=?
vx0 = 4.5 m/s
vx = 4.5 m/s
t = 1.24 s
ax = 0 m/s2
0
0
Projectile Motion Type 1
Projectile motion type 1 refers to projectiles that are horizontally
launched off an elevated surface. For example, if a ball rolls off a
table.
In this case, all the initial velocity is in the x direction, and none of it
is in the y direction.
This is the simplest type of projectile motion problem, since the
middle term of kinematics equation 2 in the y dimension goes away.
Also, all objects in free fall near earth's surface have the same
acceleration, ay = -g = -9.8 m/s2.
v0
0
24 A cannon ball is shot horizontally from a cannon at a
height of 15 m with a velocity of 20 m/s. How far
away will the cannon ball land?
A 35 m
C 60 m
D 65 m
E I need help
https://njctl.org/video/?v=fUuCkZsEims
Answer
B 50 m
24 A cannon ball is shot horizontally from a cannon at a
height of 15 m with a velocity of 20 m/s. How far
away will the cannon ball land?
A 35 m
Answer
B 50 m
C 60 m
A
D 65 m
E I need help
https://njctl.org/video/?v=fUuCkZsEims
[This object is a pull
tab]
25 A marble rolls off a table from a height of 0.8 m with
a velocity of 3 m/s. Then another marble rolls off the
same table with a velocity of 4 m/s. Which values
are the same for both marbles?
A The final speeds of the marbles.
B The time each takes to reach the ground.
C The distance from the base of the
table where each lands.
D I need help
https://njctl.org/video/?v=lSwI17d1uf0
Answer
Justify your answer qualitatively, with no equations or
calculations.
25 A marble rolls off a table from a height of 0.8 m with
a velocity of 3 m/s. Then another marble rolls off the
same table with a velocity of 4 m/s. Which values
are the same for both marbles?
Justify your answer qualitatively, with no equations or
calculations.
Answer
A The final speeds of the marbles.
B The time each takes to
reach the ground.
B
C The distance from the base of the
table where each lands.
D I need help
https://njctl.org/video/?v=lSwI17d1uf0
[This object is a pull
tab]
Projectile Motion
Type 2
https://njctl.org/video/?v=tEWHuMb4pJM
Return to
Table of
Contents
Projectile Motion Type 2
Projectile motion type 2 is when an object is launched upwards at
an angle, then lands at the same height it was launched from. For
example, kicking a soccer ball across the field.
vy
vy
v
vx
vx
v
vx = v
vy
v
vx
vx
vy
v
The analysis is similar to projectile motion type 1, except the initial
velocity now has a vertical component.
Projectile Motion Type 2
These problems typically start as "a ball is thrown at 10 m/s at an
angle of 20 degrees above the horizontal. How far away will it land?
How high will it go "
In that case, the given speed, v0, is the magnitude of the velocity
vector. It is the hypotenuse of a right triangle. We need to solve for
the sides, which represent vy0 and vx0.
v0
vy0
θ=20°
vx0
vy
vy
v
vx
vx
v
vx = v
vy
v
vx
vx
vy
v
Example 3: Projectile Motion Type 2
A ball is thrown at 10 m/s at an angle of 20 degrees above the
horizontal. How far away will it land? How high will it go?
answers on next pages
Example 3: Projectile Motion Type 2
A ball is thrown at 10 m/s at an angle of 20 degrees above the
horizontal. How far away will it land? How high will it go?
First, resolve the initial velocity into its x and y components, using
trigonometry.
/s
m
0
1
=
v
0
vy0
θ=20°
vx0
continued on next page
Example 3: Projectile Motion Type 2
A ball is thrown at 10 m/s at an angle of 20 degrees above the
horizontal. How far away will it land? How high will it go?
/s
m
10
v0 =
vy0 = 3.4 m/s
θ=20°
vx0 = 9.4 m/s
continued on next page
Example 3: Projectile Motion Type 2
A ball is thrown at 10 m/s at an angle of 20 degrees above the
horizontal. How far away will it land? How high will it go?
Compile the list of givens and calculated x and y velocity
components.
y
x
/s
m
=10
v0
θ=20°
vx0=9.4 m/s
vy0=3.4 m/s
x0 = 0 m
x=?
vx0 = 9.4 m/s
vx = 9.4 m/s
t=?
ax = 0 m/s2
y0 = 0 m
y=0m
vy0 = 3.4 m/s
vy = changes
t=?
ay = -9.8 m/s2
Do we have enough information to calculate how long the ball will
be in the air?
continued on next page
Example 3: Projectile Motion Type 2
A ball is thrown at 10 m/s at an angle of 20 degrees above the
horizontal. How far away will it land? How high will it go?
Yes! Just like with Type 1 Projectile Motion, we currently have
enough information to solve for air time. Use kinematics equation 2
in the y dimension.
y
However, unlike Type 1, the middle
term is no longer zero. Substitute
in the givens and solve for t.
continued on next page
y0 = 0 m
y=0m
vy0 = 3.4 m/s
vy = changes
t=?
ay = -9.8 m/s2
Example 3: Projectile Motion Type 2
A ball is thrown at 10 m/s at an angle of 20 degrees above the
horizontal. How far away will it land? How high will it go?
Substitute in the givens and the calculated velocity in the y
direction, vy0 = 3.4 m/s.
0
0
y
y0 = 0 m
y=0m
vy0 = 3.4 m/s
vy = changes
t=?
ay = -9.8 m/s2
continued on next page
Example 3: Projectile Motion Type 2
A ball is thrown at 10 m/s at an angle of 20 degrees above the
horizontal. How far away will it land? How high will it go?
Add time to the list of givens. Next, we can solve for how far away it
will land, x.
y
x
x0 = 0 m
x=?
vx0 = 9.4 m/s
vx = 9.4 m/s
t = 0.7 s
ax = 0 m/s2
y0 = 0 m
y=0m
vy0 = 3.4 m/s
vy = changes
t = 0.7 s
ay = -9.8 m/s2
continued on next page
Example 3: Projectile Motion Type 2
A ball is thrown at 10 m/s at an angle of 20 degrees above the
horizontal. How far away will it land? How high will it go?
Use kinematics equation 2 in the x direction to solve for x.
x
x0 = 0 m
x = 6.6 m
vx0 = 9.4 m/s
vx = 9.4 m/s
t = 0.7 s
ax = 0 m/s2
0
continued on next page
0
Example 3: Projectile Motion Type 2
A ball is thrown at 10 m/s at an angle of 20 degrees above the
horizontal. How far away will it land? How high will it go?
Do we have enough information to solve for the maximum height
that the ball will reach?
x
y
x0 = 0 m
x = 6.6 m
vx0 = 9.4 m/s
vx = 9.4 m/s
t = 0.7 s
ax = 0 m/s2
y0 = 0 m
y=0m
vy0 = 3.4 m/s
vy = changes
t = 0.7 s
ay = -9.8 m/s2
continued on next page
Example 3: Projectile Motion Type 2
A ball is thrown at 10 m/s at an angle of 20 degrees above the
horizontal. How far away will it land? How high will it go?
Yes, but a new observation is required.
The ball stops traveling upwards, it stops moving in the y direction,
then begins falling back downwards towards earth. In that instant,
there is no velocity in the y direction, but it still has velocity in the x
dimension.
At the peak of the projectile's trajectory, all the velocity is in the x
dimension, so vy-top = 0 m/s and vx-top = vx.
vx = vx
vy = 0
vy
v
vx
continued on next page
Example 3: Projectile Motion Type 2
A ball is thrown at 10 m/s at an angle of 20 degrees above the
horizontal. How far away will it land? How high will it go?
Another observation: In projectile motion type 2, the time to the top
is equal to 1/2 the total air time. It takes 0.7 s to reach the ground
again, so it takes 0.35 s to reach the very top.
vx = vx
y
y0 = 0 m
y=0m
vy0 = 3.4 m/s
vy = changes
vy-top = 0 m/s
t = 0.7 s
ttop = t/2 = 0.35 s
ay = -9.8 m/s2
vy = 0
vy
v
vx
continued on next page
Example 3: Projectile Motion Type 2
A ball is thrown at 10 m/s at an angle of 20 degrees above the
horizontal. How far away will it land? How high will it go?
Use kinematics equation 2 in the y direction to solve for the
maximum height (ytop).
y
y0 = 0 m
y=0m
vy0 = 3.4 m/s
vy = changes
vy-top = 0 m/s
t = 0.7 s
ttop = t/2 = 0.35 s
ay = -9.8 m/s2
0
Don't forget this
negative sign!
Substitute in only
the y component
of initial velocity.
Projectile Motion Type 2
There are equations for air time, horizontal range and maximum
height, but they are not worth memorizing.
They can be derived algebraically by combining kinematics
equations.
Let's calculate time in the air using the y direction equations.
Projectile Motion Type 2
There are equations for air time, horizontal range and maximum
height, but they are not worth memorizing.
They can be derived algebraically by combining kinematics
equations.
Let's calculate time in the air using the y direction equations.
Y
y0 = 0 m
ytop = 0 m
vy0 = v0sinθ
vy = changes
t=?
ay = -9.8 m/s2
v0
θ
vx0
vy0
Projectile Motion Type 2
Calculate time in the air using the y direction equations.
Y
y0 = 0 m
ytop = 0 m
vy0 = v0sinθ
vy = changes
t=?
ay = -9.8 m/s2
This equation will only
work for projectile motion
type 2. Learn to derive it.
0
0
Projectile Motion Type 2
Substituting this result into different kinematics equations
provides equations for horizontal range and maximum height:
Air time:
Horizontal range:
Maximum Height:
A
B
C
D
E
is zero.
remains constant.
continuously increases.
continuously decreases.
I need help
https://njctl.org/video/?v=44fniKpLtMs
Answer
26 Ignoring air resistance, the horizontal component
of a projectile's velocity:
26 Ignoring air resistance, the horizontal component
of a projectile's velocity:
is zero.
remains constant.
continuously increases.
continuously decreases.
I need help
Answer
A
B
C
D
E
https://njctl.org/video/?v=44fniKpLtMs
B
[This object is a pull
tab]
https://njctl.org/video/?v=f0RE3-aSLjU
Answer
27 A ball is thrown with a velocity of 20 m/s at an angle
of 60° above the horizontal. What is the horizontal
component of its instantaneous velocity at the exact
top of its trajectory?
A 10 m/s
B 17 m/s
C 20 m/s
D zero
E I need help
Answer
27 A ball is thrown with a velocity of 20 m/s at an angle
of 60° above the horizontal. What is the horizontal
component of its instantaneous velocity at the exact
top of its trajectory?
A 10 m/s
B 17 m/s
C 20 m/s
D zero
A
E I need help
[This object is a pull
tab]
https://njctl.org/video/?v=f0RE3-aSLjU
A
B
C
D
E
is zero.
remains a non-zero constant.
continuously increases.
continuously decreases.
I need help
https://njctl.org/video/?v=or6tYkO7yQk
Answer
28 Ignoring air resistance, the magnitude of thehorizontal
component of a projectile's acceleration:
28 Ignoring air resistance, the magnitude of thehorizontal
component of a projectile's acceleration:
is zero.
remains a non-zero constant.
continuously increases.
continuously decreases.
I need help
Answer
A
B
C
D
E
https://njctl.org/video/?v=or6tYkO7yQk
A
[This object is a pull
tab]
A
B
C
D
E
0°
30°
45°
60°
I need help
https://njctl.org/video/?v=mHSHtsSPraA
Answer
29 At what angle should a water-gun be aimed in
order for the water to land with the greatest
horizontal range?
29 At what angle should a water-gun be aimed in
order for the water to land with the greatest
horizontal range?
0°
30°
45°
60°
I need help
Answer
A
B
C
D
E
C
[This object is a pull
tab]
https://njctl.org/video/?v=mHSHtsSPraA
A 30° and 80°
B 20° and 70°
C 30° and 70°
D 20° and 60°
E I need help
https://njctl.org/video/?v=-NWgZcxdtbI
Answer
30 An Olympic athlete throws a javelin at six different
angles above the horizontal, each with the same speed:
20°, 30°, 40°, 60°, 70° and 80°. Which two throws
cause the javelin to land the same distance away?
30 An Olympic athlete throws a javelin at six different
angles above the horizontal, each with the same speed:
20°, 30°, 40°, 60°, 70° and 80°. Which two throws
cause the javelin to land the same distance away?
Answer
A 30° and 80°
B 20° and 70°
B
C 30° and 70°
D 20° and 60°
E I need help
[This object is a pull tab]
https://njctl.org/video/?v=-NWgZcxdtbI
A 1.4 R
B R/2
C 2R
D 4R
E I need help
https://njctl.org/video/?v=q2xmzZSJ0jw
Answer
31 You kick a soccer ball sitting on the ground and it leaves
with a velocity, v0, and an angle, θ, where 00 < θ < 900
with the ground. You then kick it again, this time with an
initial velocity 2v 0 at the same angle. The ball traveled a
distance, R, with the first kick. How far did it go after the
second kick?
31 You kick a soccer ball sitting on the ground and it leaves
with a velocity, v0, and an angle, θ, where 00 < θ < 900
with the ground. You then kick it again, this time with an
initial velocity 2v 0 at the same angle. The ball traveled a
distance, R, with the first kick. How far did it go after the
second kick?
Answer
A 1.4 R
B R/2
C 2R
D 4R
E I need help
D
[This object is a pull tab]
https://njctl.org/video/?v=q2xmzZSJ0jw
B
C
It is less than its initial speed.
It is equal to its initial speed.
D
E
It is greater than its initial speed.
I need help
https://njctl.org/video/?v=jOGkV3lyGzE
Answer
32 When a football in a field goal attempt reaches its
maximum height, how does its speed compare to its
initial speed? (Justify your answer.)
A It is zero.
32 When a football in a field goal attempt reaches its
maximum height, how does its speed compare to its
initial speed? (Justify your answer.)
A It is zero.
It is less than its initial speed.
It is equal to its initial speed.
D
E
It is greater than its initial speed.
B
I need help
Answer
B
C
https://njctl.org/video/?v=jOGkV3lyGzE
[This object is a pull
tab]
33 A stone is thrown horizontally from the top of a tower
at the same instant a ball is dropped vertically.
Which object is traveling faster when it hits the level
ground below? (Justify your answer.)
It is impossible to tell from the information given.
the stone
the ball
Neither, since both are traveling at the same speed.
I need help
https://njctl.org/video/?v=EBlb52EfXbY
Answer
A
B
C
D
E
33 A stone is thrown horizontally from the top of a tower
at the same instant a ball is dropped vertically.
Which object is traveling faster when it hits the level
ground below? (Justify your answer.)
It is impossible to tell from the information given.
the stone
the ball
Neither, since both are traveling at the same speed.
I need help
Answer
A
B
C
D
E
B
https://njctl.org/video/?v=EBlb52EfXbY
[This object is a pull
tab]
34 A plane flying horizontally at a speed of 50.0 m/s and
at an elevation of 160 m drops a package. Two
seconds later it drops a second package. How far
apart will the two packages land on the ground?
A
100 m
C 180 m
D 210 m
E
I need help
https://njctl.org/video/?v=VO-55Jzduzc
Answer
B 170 m
34 A plane flying horizontally at a speed of 50.0 m/s and
at an elevation of 160 m drops a package. Two
seconds later it drops a second package. How far
apart will the two packages land on the ground?
A
100 m
B 170 m
Answer
C 180 m
D 210 m
E
A
I need help
https://njctl.org/video/?v=VO-55Jzduzc
[This object is a pull
tab]
35 A projectile is fired with an initial speed of 30 m/s at
an angle of 30° above the horizontal.
A 2s
B 3s
C 5s
D 7s
E I need help
https://njctl.org/video/?v=tyH48qKkjyg
Answer
Determine the total time in the air.
35 A projectile is fired with an initial speed of 30 m/s at
an angle of 30° above the horizontal.
Determine the total time in the air.
Answer
A 2s
B 3s
C 5s
B
D 7s
E I need help
https://njctl.org/video/?v=tyH48qKkjyg
[This object is a pull
tab]
36 A projectile is fired with an initial speed of 30 m/s at
an angle of 30° above the horizontal.
A 8.1 m
B 9.6 m
C 11 m
D 13 m
E I need help
https://njctl.org/video/?v=kSoiajRmpHA
Answer
Determine the maximum height reached by the
projectile.
36 A projectile is fired with an initial speed of 30 m/s at
an angle of 30° above the horizontal.
Determine the maximum height reached by the
projectile.
Answer
A 8.1 m
B 9.6 m
C 11 m
D 13 m
E I need help
https://njctl.org/video/?v=kSoiajRmpHA
C
[This object is a pull
tab]
37 A projectile is fired with an initial speed of 30 m/s at
an angle of 30° above the horizontal.
A 45 m
B 61 m
C 70 m
D 78 m
E I need help
https://njctl.org/video/?v=Tw-69EQ9iw8
Answer
Determine the maximum horizontal distance
covered by the projectile.
37 A projectile is fired with an initial speed of 30 m/s at
an angle of 30° above the horizontal.
Determine the maximum horizontal distance
covered by the projectile.
Answer
A 45 m
B 61 m
C 70 m
D 78 m
D
E I need help
[This object is a pull
tab]
https://njctl.org/video/?v=Tw-69EQ9iw8
38 A projectile is fired with an initial speed of 30 m/s at
an angle of 30° above the horizontal.
Determine the angle of the velocity of the projectile,
relative to the x axis, 2.0 s after firing.
B 11o below the x axis
C 14o above the x axis
D 19o above the x axis
E I need help
https://njctl.org/video/?v=3iXp_4q5QzM
Answer
A 5.4o below the x axis
38 A projectile is fired with an initial speed of 30 m/s at
an angle of 30° above the horizontal.
Determine the angle of the velocity of the projectile,
relative to the x axis, 2.0 s after firing.
A 5.4o below the x axis
Answer
B 11o below the x axis
C 14o above the x axis B
D 19o above the x axis
E I need help
https://njctl.org/video/?v=3iXp_4q5QzM
[This object is a pull
tab]
Projectile Motion
Type 3
https://njctl.org/video/?v=ZlZBf_IFaAI
Return to
Table of
Contents
Projectile Motion Type 3
The third type of
projectile motion
involves an angled
launch and a different
landing height.
The analysis is similar
to the previous types,
except when finding the
total time in the air.
Start by applying the
second kinematics
equation in the y
direction to find the
time.
vy
vy
v
v vx = v
vx
vx
vy
v
vx
vx
vy
v
vx
vy
v
Projectile Motion Type 3
For this motion, the time
to reach the peak of the
trajectory is not half the
total time in the air.
The peak time is half
the time it takes to get
to point A. The speed at
point A is equal to the
launch speed.
At point B, the projectile
is traveling faster than it
started. The horizontal
distance traveled at
point B is greater than
at point A.
vy
vy
v
v vx = v
vx
vx
vx
vy
v
A
vx
vy
v
vx
vy
v
B
Projectile Motion Type 3
The quadratic equation is required since the projectile lands
at a different height then where it started, y ≠ y 0. Set the
landing height at y = 0, and y 0 will be greater than zero.
y
y0 > 0 m
y=0m
vy0 = v0sinθ
vy = changes
t=?
ay = -g
0
Since this term is not zero, the
quadratic equation is required to
solve for t.
When using the quadratic
equation, it is not necessary to
keep track of the units until the
end of the solution.
Example 4: Projectile Motion Type 3
A ball is thrown off a 50 m tall cliff with a speed of 20 m/s at an
angle of 30 o above the horizontal. How long will it take for the
ball to hit the water below? How far will it go?
answers on next pages
Example 4: Projectile Motion Type 3
A ball is thrown off a 50 m tall cliff with a speed of 20 m/s at an
angle of 30 o above the horizontal. How long will it take for the
ball to hit the water below? How far will it go?
y
First, calculate the y component of the initial velocity.
y0 = 50 m
y=0m
vy0 = 10 m/s
t=?
ay = -g
continued on next page
Example 4: Projectile Motion Type 3
A ball is thrown off a 50 m tall cliff with a speed of 20 m/s at an
angle of 30 o above the horizontal. How long will it take for the
ball to hit the water below? How far will it go?
y
Calculate the air time.
y0 = 50 m
y=0m
vy0 = 10 m/s
t=?
ay = -g
0
Quadratic Formula
a: -4.9
b: 10
c: 50
Since negative time doesn't make sense
in this context, the answer is 4.4 s.
continued on next page
Example 4: Projectile Motion Type 3
A ball is thrown off a 50 m tall cliff with a speed of 20 m/s at an
angle of 30 o above the horizontal. How long will it take for the
ball to hit the water below? How far will it go?
vx = v
Actually, with a little
imagination, t = -2.3 s does
have a meaning!
vy
B
It is the time it would take a
projectile launched at point A
to reach point B.
Add the magnitude of that
time to the time solution from
point B to point C, and its a
projectile motion type 2
problem.
v
vx
vx
vy
v
vx
vy
A
continued on next page
v
C
Example 4: Projectile Motion Type 3
A ball is thrown off a 50 m tall cliff with a speed of 20 m/s at an
angle of 30 o above the horizontal. How long will it take for the
ball to hit the water below? How far will it go?
x
x0 = 0 m
x=?m
vx0 = 17 m/s
t = 4.4 s
ax = 0 m/s2
y
y0 = 50 m
y=0m
vy0 = 10 m/s
t = 4.4 s
ay = -9.8 m/s2
Back to the problem. Calculate
the x component of the initial
velocity:
Example 4: Projectile Motion Type 3
A ball is thrown off a 50 m tall cliff with a speed of 20 m/s at an
angle of 30 o above the horizontal. How long will it take for the
ball to hit the water below? How far will it go?
x
x0 = 0 m
x=?m
vx0 = 17 m/s
t = 4.4 s
ax = 0 m/s2
y
y0 = 50 m
y=0m
vy0 = 10 m/s
t = 4.4 s
ay = -9.8 m/s2
Finally, calculate horizontal
range:
0
0
39 A projectile is fired from the edge of a cliff 200 m high
with an initial speed of 30 m/s at an angle of 45° above
the horizontal.
Determine the total time in the air.
B 5.5 s
C 8.9 s
D 11 s
E I need help
https://njctl.org/video/?v=79A0al4cOm8
Answer
A 3.9 s
39 A projectile is fired from the edge of a cliff 200 m high
with an initial speed of 30 m/s at an angle of 45° above
the horizontal.
Determine the total time in the air.
A 3.9 s
Answer
B 5.5 s
C 8.9 s
C
D 11 s
E I need help
https://njctl.org/video/?v=79A0al4cOm8
[This object is a pull
tab]
40 A projectile is fired from the edge of a cliff 200 m high
with an initial speed of 30 m/s at an angle of 45° above
the horizontal.
A 120 m
B 187 m
C 200 m
D 250 m
E I need help
https://njctl.org/video/?v=ln1SvI7XEvk
Answer
Determine the maximum horizontal range.
40 A projectile is fired from the edge of a cliff 200 m high
with an initial speed of 30 m/s at an angle of 45° above
the horizontal.
Determine the maximum horizontal range.
Answer
A 120 m
B 187 m
C 200 m
B
D 250 m
E I need help
https://njctl.org/video/?v=ln1SvI7XEvk
[This object is a pull
tab]
41 A projectile is fired from the edge of a cliff 200 m high
with an initial speed of 30 m/s at an angle of 45° above
the horizontal.
Determine the magnitude of the velocity just before
impact.
A 45 m/s
C 77 m/s
D 90 m/s
E I need help
https://njctl.org/video/?v=m9tBoeE1umE
Answer
B 70 m/s
41 A projectile is fired from the edge of a cliff 200 m high
with an initial speed of 30 m/s at an angle of 45° above
the horizontal.
Determine the magnitude of the velocity just before
impact.
A 45 m/s
Answer
B 70 m/s
C 77 m/s
B
D 90 m/s
E I need help
https://njctl.org/video/?v=m9tBoeE1umE
[This object is a pull
tab]
42 A projectile is fired from the edge of a cliff 200 m high
with an initial speed of 30 m/s at an angle of 45° above
the horizontal.
B -44o
C -72o
D -80o
E I need help
https://njctl.org/video/?v=INAZs-LOp8I
Answer
Determine the angle the velocity makes with the
horizontal just before impact.
A -30o
42 A projectile is fired from the edge of a cliff 200 m high
with an initial speed of 30 m/s at an angle of 45° above
the horizontal.
Determine the angle the velocity makes with the
horizontal just before impact.
A -30o
Answer
B -44o
C
C -72o
D -80o
E I need help
https://njctl.org/video/?v=INAZs-LOp8I
[This object is a pull
tab]
Calculating Landing Velocity
https://njctl.org/video/?v=NNvWCQDzDFk
Return to
Table of
Contents
Landing Velocity
To calculate landing velocity, let's zoom in on the object as it
lands. The landing velocity is a vector on an angle, composed of
x and y components. It is the hypotenuse of a right triangle.
We will find this hypotenuse using the Pythagorean theorem.
vx
vx
θ
v
vy
v
vx stays constant the entire flight.
vy
Calculate vy, using either
kinematics equation 1 or 3.
Then, use the Pythagorean
theorem to calculate the
hypotenuse.
Example 5: Landing Velocity
A ball is launched horizontally off a 30 m tall cliff with a speed of
50 m/s. Calculate the landing velocity.
Landing velocity information:
vx = 50 m/s
vy = ?
For projectile motion type 1,
there is no initial velocity in the y
dimension. It will gain speed
vertically at the same rate as an
object dropped from the same
height.
How fast is an object moving
after being dropped from rest
from 30 m high?
vx
vx
θ
v
vy
v
vy
Example 5: Landing Velocity
A ball is launched horizontally off a 30 m tall cliff with a speed of
50 m/s. Calculate the landing velocity.
Landing velocity information:
vx = 50 m/s
vy = -17 m/s
Kinematics Equation 3:
0
vy2 = vy02 + 2ay(y - y0)
vy2 = 2(-9.8)(-30) = 294
vy = -24.2 m/s
take
square
root
vx
vx
θ
v
vy
v
vy
The velocity is negative
because the vector is pointing
downwards.
Example 5: Landing Velocity
A ball is launched horizontally off a 30 m tall cliff with a speed of
50 m/s. Calculate the landing velocity.
Landing velocity information:
vx = 50 m/s
y
vy = -24.2 m/s
50
x
Magnitude:
c2 = a2 + b2
c2 = (50)2 + (-24.2)2
θ
v
vx
vy
v
-24.2
c = 55.5
v = 55.5 m/s
Direction:
θ = tan-1(-24.2/50)
θ = 26° below the
horizontal
43 If a projectile has a final horizontal velocity of 15 m/s,
and a final vertical velocity of -10 m/s, what is the
magnitude of the object's landing velocity?
A 5 m/s
C 18 m/s
D 25 m/s
E I need help
https://njctl.org/video/?v=fh-8tfEHxi4
Answer
B 12 m/s
43 If a projectile has a final horizontal velocity of 15 m/s,
and a final vertical velocity of -10 m/s, what is the
magnitude of the object's landing velocity?
A 5 m/s
B 12 m/s
Answer
C 18 m/s
D 25 m/s
C
E I need help
[This object is a pull tab]
https://njctl.org/video/?v=fh-8tfEHxi4
44 If a projectile has a final horizontal velocity of 15 m/s,
and a final vertical velocity of -10 m/s, what is the
direction of the object's landing velocity?
A 34° below the horizontal
C 34° above the horizontal
D 56° above the horizontal
E I need help
https://njctl.org/video/?v=MehwM24eJx4
Answer
B 56° below the horizontal
44 If a projectile has a final horizontal velocity of 15 m/s,
and a final vertical velocity of -10 m/s, what is the
direction of the object's landing velocity?
A 34° below the horizontal
Answer
B 56° below the horizontal
C 34° above the horizontal
A
D 56° above the horizontal
E I need help
[This object is a pull tab]
https://njctl.org/video/?v=MehwM24eJx4