Kinematics in Two
Dimensions
Section 1:
Adding Vectors Graphically
• Adding Vectors Graphically
• Remember vectors have magnitude
(length) and direction.
• When you add vectors you must maintain
both magnitude and direction
• This information is represented by an
arrow (vector)
• A vector has a magnitude and a direction
– The length of a drawn vector represents
magnitude.
– The arrow represents the direction
Larger Vector
Smaller Vector
Graphical Representation of Vectors
• Given Vector a:
Draw 2a
Draw -a
Problem set 1:
a
b
c
1. Which vector has the largest magnitude?
2. What would -b look like?
3. What would 2 c look like?
Vectors
a
b
• Three vectors
c
a
b
c
• When adding vectors graphically, align the vectors headto-tail.
• This means draw the vectors in order, matching up the
point of one arrow with the end of the next, indicating the
overall direction heading.
• Ex. a + c
• The starting point is called the origin
c
a
origin
a
b
c
• When all of the vectors have been
connected, draw one straight arrow from
origin to finish. This arrow is called the
resultant vector.
c
a
origin
a
b
• Ex.1
Draw a + b
c
a
b
• Ex.1
Draw a + b
Resultant
origin
c
a
b
• Ex. 2
Draw a + b + c
c
a
b
• Ex. 2
Draw a + b + c
Resultant
origin
c
a
b
• Ex. 3
Draw 2a – b – 2c
c
a
b
• Ex. 3
Draw 2a – b – 2c
origin
Resultant
c
Section 2: How do you name vector
directions?
Vector Direction Naming
• How many degrees is this?
N
W
E
S
Vector Direction Naming
• How many degrees is this?
N
90º
W
E
S
Vector Direction Naming
• What is the difference between 15º North
of East and 15 º East of North?
N
W
E
S
Vector Direction Naming
• What is the difference between 15º North
of East and 15º East of North? (can you
tell now?)
N
W
N
E
S
15º North of East
W
E
S
15º East of North
Vector Direction Naming
N
15º
W
S
15º North of what?
Vector Direction Naming
N
15º
E
W
S
15º North of East
15º
W
E
S
15º East of What?
N
W
15º
E
S
15º East of North
N of ___
E
___
This is the baseline.
It is the direction you
look at first
This is the direction you
go from the baseline to
draw your angle
Describing directions
• 30º North of East
– East first then 30º North
• 40º South of East
– East first then 30º South
• 25º North of West
– West first then 30º North
• 30º South of West
– West first then 30º South
Problem Set #2 (Name the angles)
30º
45º
20º
30º
20º
Intro: Get out your notes
a
b
c
1. Draw the resultant of
a–b+c
2. What would you label following angles
28º
a.
b.
18º
3. Draw the direction 15º S of W
Section 3: How do you add
vectors mathematically (not
projectile motion)
The Useful Right Triangle
• Sketch a right triangle and label its sides
c: hypotenuse
a: opposite
Ө
b: adjacent
The angle
• The opposite (a) and adjacent (b) change based
on the location of the angle in question
• The hypotenuse is always the longest side
c: hypotenuse
Ө
b: adjacent
a: opposite
• The opposite (a) and adjacent (b) change based
on the location of the angle in question
• The hypotenuse is always the longest side
c: hypotenuse
Ө
b: adjacent
a: opposite
To figure out any side when given two other sides
• Use Pythagorean Theorem
a2 + b2 = c2
c: hypotenuse
a: opposite
Ө
b: adjacent
The angle
Sometimes you need to use trig functions
c: hypotenuse
a: opposite
Ө
a: adjacent
Opp
Sin Ө = _____
Hyp
Opp
Tan Ө = _____
Adj
Adj
Cos Ө = _____
Hyp
Sometimes you need to use trig functions
c: hypotenuse
a: opposite
Ө
a: adjacent
Opp
Sin Ө = _____
Hyp
SOH CAH TOA
Opp
Tan Ө = _____
Adj
Adj
Cos Ө = _____
Hyp
More used versions
Opp
Sin Ө = _____
Hyp
Opp = (Sin Ө)(Hyp)
Adj
Cos Ө = _____
Hyp
Adj = (Cos Ө)(Hyp)
Opp
Tan Ө = _____
Adj
Ө=
Tan-1
Opp
_____
Adj
• To resolve a vector means to break it
down into its X and Y components.
Example: 85 m
25º N of W
• Start by drawing the angle
25º
• To resolve a vector means to break it
down into its X and Y components.
Example: 85 m
25º N of W
• Start by drawing the angle
• The magnitude given is always the hypotenuse
85 m
25º
• To resolve a vector means to break it down
into its X and Y components.
Example: 85 m
25º N of W
• this hypotenuse is made up of a X component (West)
• and a Y component (North)
85 m
North
25º
West
In other words:
I can go so far west along the X axis and so
far north along the Y axis and end up in the
same place
finish
finish
85 m
North
origin
West
25º
origin
• If the question asks for the West
component: Solve for that side
– Here the west is the adjacent side
Adj = (Cos Θ)(Hyp)
85 m
25º
West
or Adj.
• If the question asks for the West
component: Solve for that side
– Here the west is the adjacent side
Adj = (Cos Θ)(Hyp)
Adj = (Cos 25º)(85) = 77 m W
85 m
25º
West
or Adj.
• If the question asks for the North
component: Solve for that side
– Here the north is the opposite side
Opp = (Sin Θ)(Hyp)
North
or
Opp.
85 m
25º
• If the question asks for the North
component: Solve for that side
– Here the west is the opposite side
Opp = (Sin Θ)(Hyp)
Opp = (Sin 25º)(85) = 36 m N
North
or
Opp
85 m
25º
Resolving Vectors Into Components
• Ex 4a. Find the west component of 45 m
19º S of W
Resolving Vectors Into Components
• Ex 4a. Find the west component of 45 m
19º S of W
• Ex 4a. Find the south component of 45 m
19º S of W
• Ex 4a. Find the south component of 45 m
19º S of W
Remember the wording.
These vectors are at right
angles to each other.
5 m/s
forward
5 m/s
Redraw and
it becomes
velocity = 30 m/s
down
Right
angle
Hypotenuse
= Resultant
speed
30 m/s
Section 4 (Solving for a resultant)
• Ex. 6
Find the resultant of 35.0 m, N
and 10.6 m, E.
• Start by drawing a vector diagram
• Then draw the resultant arrow
• Ex. 6
Find the resultant of 35.0 m, N
and 10.6 m, E.
Then draw the resultant vector and angle
– The angle you find is in the triangle closest to the origin
• Now we use Pythagorean theorem to figure out
the resultant (hypotenuse)
• Then inverse tangent to figure out the angle
• The answer needs a magnitude, angle, and
direction
Problem Set 3: Resolve the following
vectors
1) 48m, S and 25m, W
2) 12.5m, S and 78m, N
Problem Set #3
1) 48m, S and 25m, W
Section 4: How does projectile
motion differ from 2D motion
(without gravity)?
Projectile Motion
• Projectile- Object that is launched by a
force and continues to move by its own
inertia
• Trajectory- parabolic path of a projectile
• Projectile motion involves an object
moving in 2D (horizontally and vertically)
but only vertically is influenced by gravity.
• The X and Y components act
independently from each other and will be
separated in our calculations.
X and Y are independent
• X axis has uniform motion since gravity
does not act upon it.
X and Y are Independent
• Y axis will be accelerated by gravity -9.8
m/s2
The equations for uniform acceleration, from
unit one, can be written for either x or y
variables:
• If we push the ball harder, giving it a
greater horizontal velocity as it rolls off the
table, the ball would take _________ time
to fall to the floor.
Horizontal and vertical movement
is independent
• If we push the ball harder, giving it a
greater horizontal velocity as it rolls off the
table, the ball would:
– Y axis: take the same time to fall to the floor.
– X axis: It would just go further.
Solving Simple Projectile Motion Problems
• You will have only enough information to deal with the y
or x axis first
• You cannot use the Pythagorean theorem since X and
Y-axes are independent
• Time will be the key: The time it took to fall is the same
time the object traveled vertically.
• dx = (vx)(t)
motion.
is the equation for the horizontal uniform
• If you don’t have 2 of three x variable you will have to
solve for t using gravity and the y axis
Equations Solving Simple Projectile
Motion Problems
• Do not mix up y and x variables
• dy – height
(this is negative if falling down)
• dx – range (displacement x)
For all projectile motion problems
• Draw a diagram
• Separate the X and Y givens
• Something is falling in these problems
X Givens
dX =
vX =
t=
Y Givens
a = -9.8 m/s
…
…
…
Example Problem 8
• A stone is thrown horizontally at 7.50 m/s
from a cliff that is 68.4 m high. How far
from the base of the cliff does the stone
land?
Write out your x and y givens separately
• A stone is thrown horizontally at 7.50 /s
from a cliff that is 68.4 m high. How far
from the base of the cliff does the stone
land?
X givens
Y givens
• A stone is thrown horizontally at 7.50 m/s
from a cliff that is 68.4 m high. How far
from the base of the cliff does the stone
land?
X givens
Y givens
Ex. 9
A baseball is thrown horizontally with a
velocity of 44 m/s. It travels a horizontal
distance of 18, to the plate before it is
caught.
a) How long does the ball stay in the air?
b) How far does it drop during its flight?
• A baseball is thrown horizontally with a
velocity of 44 m/s. It travels a horizontal
distance of 18, to the plate before it is
caught.
– How long does the ball stay in the air?
– How far does it drop during its flight?
X givens
Y givens
• A baseball is thrown horizontally with a
velocity of 44 m/s. It travels a horizontal
distance of 18, to the plate before it is
caught.
– How long does the ball stay in the air?
– How far does it drop during its flight?
X givens
Y givens
• A baseball is thrown horizontally with a
velocity of 44 m/s. It travels a horizontal
distance of 18, to the plate before it is
caught.
– How long does the ball stay in the air?
– How far does it drop during its flight?
X givens
Y givens
Example
10
1. What is the initial vertical velocity of the ball?
voY = 0 m/s
Same as if it was dropped from rest
2. How much time is required to get to the
ground?
Since voY = 0 m/s use
2(-10)
-10
t = 1.4 s
3. What is the vertical acceleration of the ball at
point A?
aoY = -10 m/s2
always
4. What is the vertical acceleration at point B?
aoY = -10 m/s2
always
5. What is the horizontal velocity of the ball at
point C?
vX = 5 m/s
(does not change)
6. How far from the edge of the cliff does the ball
land in the x plane?
X givens
vX = 5 m/s
dx = (vX)(t)
t = 1.4
dx = (5)(1.4) = 7m
dx = ?
• What will happen if drops a package when
the plane is directly over the target?
• The package has the same horizontal
velocity as the plane and would land far
away from the target.
Section 5: What do you do
different if you have projectile
motion and V0Y is not equal to 0
Projectile Motion Concepts
Arrows represent x and y velocities (g always = 10 m/s2 down)
Key points in a projectiles path
VoY = 0 m/s
• When a projectile is at its highest point its vfy = 0.
This means it stopped moving up.
• Use vfy = 0 in a question that asks you to predict
the vertical distance (how high)
Key points in a projectiles path
• If an object lands at the same height its
vertical velocities final magnitude equals
its initial but is in the opposite direction
(down)
VoY = +30 m/s
VfY = -30 m/s
• The time it takes to rise to the top equals the
time it takes to fall.
– Givens to use to find time to the top:
VoY = +30 m/s
VfY = 0 m/s
– Givens to use to find time of entire flight:
VoY = +30 m/s
VfY = -30 m/s
VoY = +30 m/s
VfY = -30 m/s
Key points in a projectiles path
• If a projectile lands
below where it is
launched the vfy
magnitude will be
greater than voy and in
the reverse direction
Ex. 11 A ball of m = 2kg is thrown from the
ground with a horizontal velocity of 5 m/s
and rises to a height of 45 m.
1. What happens to velocity in the x direction?
Why?
It stays constant during the entire flight
(no forces acting in the x direction)
2. What happens to velocity in the y direction?
Why?
It accelerates (the force of gravity is
pulling it to Earth)
3. Where is the projectile traveling the fastest?
Why?
A and E (has the largest V component)
Y
4. Where is the projectile traveling the slowest?
What is its speed at this point?
C
(has only VX component
VY=0)
5. Where is the acceleration of the projectile the
greatest? Why? All (g stays -10m/s2)
6. What is the acceleration due to gravity at point
B?
2
All
(g stays -10m/s )
7. What is the initial vertical velocity the ball is
thrown with? Must solve
vf2 = vo2 + 2ad
aY = -10m/s2
d = 45m
vo = ?
Vf = 0
vo = √(vf2 – 2ad)
vo = √(02 – 2(-10)(45)
vo = 30 m/s up
VoY = +30 m/s
VfY = -30 m/s
8. What is the time required to reach point C if
thrown from the ground?
Must solve
Y givens
aY = -10m/s2
vf = vo + at
vo = +30 m/s
t = (vf – vo)
Vf = 0 m/s
t=?
t = (0 – 30)
a
-10
t=3s
9. From point C, what is the time needed to reach
the ground?
Same as time it
took to get to the
top
t=3s
10. What is the horizontal velocity at point A?
5 m/s (never changes horizontally while in
the air)
11. What is the horizontal acceleration of the ball
at point E?
ax = 0 m/s2 (they asked for acceleration
no horizontal acceleration)
vx stays 5 m/s
12. What is the vertical acceleration due to gravity
at point E?
2
aY = -10 m/s
13. How far in the x plane (what is the range) does
the ball travel?
Must solve
X givens
dX = (vX)(t)
t= 6 seconds total in air
vX = 5 m/s
dX = ?
dX = (5)(6) = 30 m
14. What would happen to the problem if the
objects mass was 16 kg
Nothing would change. The acceleration
due to gravity is the same for any mass
• More complex projectile motion problems
require you separate a resultant velocity
vector into its components using soh-cahtoa
• A stone is thrown at 25 m/s at a 40º angle
with the horizon. Start with the finding the
vx and voy
• Then solve the problem like we have
voy
Example
• The punter on a football team tries to kick
a football with an initial velocity of 25.0 m/s
at an angle of 60.0º above the ground,
what range (dx) does it travel?
Example
• The punter on a football team tries to kick
a football with an initial velocity of 25.0 m/s
at an angle of 60.0º above the ground,
what range (dx) does it travel?
Example
•
The punter on a football team tries to kick a football with an initial velocity of
25.0 m/s at an angle of 60.0º above the ground, what range (dx) does it
travel?
•
The punter on a football team tries to kick a football with an initial velocity of
25.0 m/s at an angle of 60.0º above the ground, what range (dx) does it
travel?
•
The punter on a football team tries to kick a football with an initial velocity of
25.0 m/s at an angle of 60.0º above the ground, what range (dx) does it
travel?
45º will get you the greatest range
• Range is dx
• Horizontal displacement
Besides 45º, two sister angles will give
you the same range
• 45º is would give you the
greatest dx
• Any similar degree variation on
either side of 45º would give
you the same dx
• Ex these would give you the
same dx.
• 40º and 50º
• 30º and 60º
• 15º would give you the same
range as what? ___________
Classwork/Homework
• 2D motion Packet
• Pg 2 Exercise 10-16
• Honors Addition:
• Book Pg 79 #16,17,18,20,22,27,31
• Try 35