Chapter 3 Projectile motion
Kinematics in
Dimensions
Projectile Motion
• Projectile motion is motion in two directions
• Motion in the x-direction is independent of
the y-direction
• Motion in the y-direction is independent of
the x-direction
What is a Projectile?
• A projectile is any object that is placed into
free flight and is being affected by gravity.
g  9.8 m s 2
• The path of a projectile is called the
trajectory.
The Full Parabola
• The key to the full parabola is symmetry.
• Try to identify some points of symmetry.
Throw
Vy
Vx
Half Parabola Timing
• The time of flight of a half parabolic
path is equal to that of simply
dropping the object from the same
height.
Horizontal velocity (vx) has no
affect on flight time because it
is not affected by gravity.
Jill drops the yellow ball and
throws the red ball horizontally.
Which ball will hit the ground
first?
X & Y are Independent
Object launched horizontally
Multi media studio
What do you notice about the horizontal
velocity in each of the following
animations?
• Horizontally launched projectiles
• The plane and the package
• The truck and the ball
A stone is thrown
horizontally from the top
of a 78.4m high cliff at
5m/s.
Var
a
v1
v2
d1
a) How long does it
take to reach the
bottom?
X
Y
Want
d2
t
v  v  2 gd
2
2
2
1
v2  v1  gt
1 2
d  v1t  gt
2
d  (v1  v2 )t
A stone is thrown horizontally from the top of a
78.4m high cliff at 5m/s.
a) How long does it take to reach the bottom?
1 2
d y = v y t + at
2
2y
t
g
2(78.4m)
t
9.8 m s2
t  4.0s
A stone is thrown horizontally from the top of a
78.4m high cliff at 5m/s.
b) How far from the base does it land?
d x = vx t
dx  5
dx
m
s
(4.0s)
d x  20m
A stone is thrown horizontally from the top of a
78.4m high cliff at 5m/s.
c) What are the final vy and vx ?
vy 2  v
2
2
y1
 2ad
vy 2  v
2
y1
vx 2  vx1
 2ad
vy 2   (0 s )  2(9.8 m s2 )(78.4m)
m
2
vy 2  39.2m / s
Sample Partial WS 7a #1
• A cannon nestled in the side of a cliff (d1y = 65m)
fires a cannon ball at 26 ms . How long until the ball
splashes into the sea?
Fire
Sample Problem
• A toy car is raced off a table (1.1m high)
onto the floor below.
– How long did it take for the car to crash on the floor?
0
d2 y  d1y  v1yt  12 gt 2
t
t
2  d 2 y  d1 y 
g
2  0m  1.1m 
9.8 sm2
t  0.474 s
Projectile Motion Type
Not all object are
launch horizontally
Objects can be
launched at an
angle
• Recall the trajectory of the golf ball when hit with
a 3 iron.
• What would the trajectory of a 9 iron look like?
• The loft of the club changed the launch angle.
Projectile Motion Path Plot
Vertical Position dy (m)
30
25
20
15
10
5
0
0
10
20
30
40
50
Horizontal Position dx (m)
• Object 1 was launched at 60o
• Object 2 was launched at 30o
60
Projectile Motion Path Plot
Vertical Position dy (m)
30
25
20
15
10
5
0
0
10
20
30
40
50
60
Horizontal Position dx (m)
• Object 1 was launched from a 25m
high cliff at 0o
• Object 2 was launched at 60o
Initial Velocity Breakdown
• When an object is launched at some angle, it’s initial
velocity (v1) can be broken down into two components.
– Horizontal Component (Vx)
– Vertical Component (Vy)
• What shape is formed?
• Consider also the launch angle (q).
v1 horizontal and vertical
v1yPlease Note:
components are independent of one
another. The only commonality is time.
q
v1x
Initial Velocity Breakdown (Cont.)
• Consider the breakdown from the previous slide again.
• There are trigonometric relationships between the sides
and angles of a right triangle.
Important!
v1
q
v1x
v1y
v1 y
opp
sin q 
hyp
sin q 
adj
cos q 
hyp
v1x
cos q 
v1
opp
tan q 
adj
v1 y
tan q 
v1
v1x
Practicing Trig Functions
• Consider the triangle below.
• Solve for the unknown values.
13
Searching for x
Searching for y
adj
cos q 
hyp
opp
sin q 
hyp
y cos  22.62   x
13
22.62°
x
y
sin  22.62  
13
x  13 cos  22.62 y  13 sin  22.62
x  12
y5
Sample Velocity Breakdown
• A dart gun is fired at an angle of 30° with a
muzzle velocity of 40m/s.
• Calculate the components of the velocity?
v1
q
v1x
v1x  34.6 ms
v1y  20 ms
Horizontal
Vertical Component (y)
(x)
v1y
vv11yx
cosqq 
sin
vv11
v11yx  v1 sin
cosqq
cos30
30
v11yx   40 ms  sin
m m
v11yx  20
34.6
s s
• Given an initial velocity of 40m/s and an angle of 25
• find v1x & v1y
Searching for y
Searching for x
O
sin q 
H
A
cos q 
H
sin q 
vx
cos q 
v
vx  40cos 25
o
vx  36.25
vy
v
vy  40sin 25
o
vy  16.9
v1y
q
=25o
v1x
Sample Full Parabola Problem
• A golf ball is struck at an angle of q = 36° with
the horizontal at a velocity of 45m/s.
– What are the components of the velocity (v1x and v1y)?
Horizontal Component
Strike
v1x  v1 cos q
v1x   45 ms  cos 36
v1x  36.4 ms
v1
q
v1x
v1y
Vertical Component
v1 y  v1 sin q
v1y   45 ms  sin 36
v1y  26.4 ms
Problem Solving Strategies
•
Solve for the horizontal component Vxi
–
•
Use trig functions
Solve for the vertical component Vyi
–
Use trig functions
•
Solve each direction (x & y) separately
•
Symmetry can be used when the launching &
landing places are the same height.
A football player kicks a ball at 27m/s at
an angle of 30°.
Horizontal
a) Find the hang time
b) find the horizontal
distance the ball
travels.
b) The maximum
height of the ball.
V
V
K
a
g
-9.8
V1x
v1y
v2x
v2y
dx
t
K
W
Vertical
dx
W
dy
dy
t
t
Step 1: Solve for the horizontal and vertical components
(V1x & V1y )
30 o
Searching for y
Searching for x
vy  v sin q
vx  v cosq
vx  27cos30
o
vx  23.4
V1x= ?m/s
m
s
vy  27sin30
vy  13.5
m
s
V1y= ?m/s
Problem Solving Strategies
Problem Solving Strategies
Step 2: Solve each direction (x & y) separately
Symmetry can be used when the launching & landing
places are the same height.
15.0m/s
12.5m/s
10.0m/s
7.50m/s
5.00m/s
2.50m/s
0.00m/s
A football player kicks a ball at 27m/s at
an angle of 30°.
a) Find the hang time
b) find the horizontal
distance the ball
travels.
b) The maximum height
of the ball.
Horizontal
V
K
W
a
Vertical
V
K
g
-9.8
v1x
23.4
V1y 13.5
v2x
23.4
v2 y
dx
t
dx
W
dy
dy
t
t
A football player kicks a ball at 27m/s at
an angle of 30°.
a) Find the hang time
v2  v1
v2  v1  gt
t
v2 y  v y1
g
(v1 )  v1  gt
13.5  13.5
t
9.8 sm2
t  2.75 s
m
s
m
s
A football player kicks a ball at 27m/s at
an angle of 30°.
b) Find the horizontal distance
dx
vx 
t
d x  (23.4 )*(2.75s)
m
s
d x  64.3m
A football player kicks a ball at 27m/s at
an angle of 30°.
c) Find the maximum height
What is true about the vertical velocity at the maximum height?
15.0m/s
12.5m/s
10.0m/s
7.50m/s
5.00m/s
2.50m/s
0.00m/s
A football player kicks a ball at 27m/s at an angle
of 30°. Find the max height
v v
d
2g
2
2
v  v  2gd
2
2
2
1
0  (13.5 )
d
2( 9.8 )
2
m 2
s
m
s2
d  9.3m
2
1
•
•
•
An arrow is shot at 44m/s at an angle of 60°
Find the maximum height of the arrow.
Find the horizontal distance the arrows
travels.
Find the hang time
@ d y (max) vy  0 ms
dX
@ dy  0
 v2 y  v1 y
Step 1: Solve for the horizontal and vertical components
(V1x & V1y )
60 o
Searching for y
Searching for x
vx
cos q 
v
sin q 
vx  44cos60
o
vx  22
m
s
V1x= ?m/s
vy
v
vy  44sin 60
vy  38.1
m
s
V1y= ?m/s
Problem Solving Strategies
An arrow is shot at 44m/s at an angle of 60°
a) Find the hang time
Horizontal
V
b) find the horizontal
distance the arrows
travels.
K
W
a
v1x
22
dx
t
V
K
g
-9.8
W
v1y 38.1
v2x
c) The maximum
height of the arrow.
Vertical
v2y
dx
dy
dy
t
t
A football player kicks a ball at 44m/s at
an angle of 60°.
a) Find the hang time
vy 2  vy1
vy 2  vy1  gt
t
v2 y  v y1
g
(vy1 )  vy1  gt
38.1  38.1
t
9.8 sm2
m
s
t  7.78 s
m
s
A football player kicks a ball at 44m/s at
an angle of 60°.
b) Find the horizontal distance
dx
vx 
t
d x  (22m / s)*(7.78s)
d x  171.2m
A football player kicks a ball at 44m/s at an angle
of 60°. Find the max height
Recall, vy=0 at dy max
v  v  2ad
2
y2
2
y1
d
0  (38.1 )
d
2( 9.8 )
2
m 2
s
m
s2
d  74.0m
v v
2
y2
2a
2
y1
Adding Vectors Graphically
• You walk 5m @ 0o and then turns to walk 6m @90o.
Finally, you turn to walk 8 m at 200°. What is your
displacement?
Addition is
commutative!
dR  4@ 120
Adding Force Vectors Analytically
cos q 
opposite
sin q 
Opposite
Hypontenuse
q
adjacent
Hypontenuse
adjacent
tan q 
opposite
adjacent
o
tan ( )  q
a
1
Components of Vectors
Finding the vector magnitude and direction
when you know the components.
A  Ax2  Ay2
tan q 
Ay
Ax
 q  tan
1
Ay
Ax
Recall: q is measured
from the positive x axis.
Caution: Beware of the tangent function.
Always consider in which quadrant the vector lies when
dealing with the tangent function.
I
II
q  tan 1 (5/  8.66)
q  30
q  150
q  tan 1 (5/ 8.66)
q  30
5
5
-8.66
-8.66
8.66
-5
-5
q  tan 1 (5/  8.66)
q  30
q  210
III
8.66
q  tan 1 (5/ 8.66)
q  30
q  330
IV
Adding Vectors Analytically
Ax  A cosq
Ay  A sin q
A  4.5 N @ 30
By  B sin q
B  7 N @ 210
Bx  B cosq
Cy  C sin q
C  6 N @150 Cx  C cos q

o

o

o
Magnitude
Angle
4.5N
7N
6N
---------------
30o
210o
150o
------------
X component Y component
3.89N
-6.06N
-5.19N
Rx=-7.36N
2.25N
-3.5N
3.0N
Ry=1.7N
Adding Vectors Analytically
Rmag  R  R
2
x
q  tan (
1
2
y
Rmag  (7.36)  1.7
2
2
Magnitude
Angle
4.5N
7N
6N
---------------
30o
210o
150o
------------
R=7.55N
Ry
)
Rx
1.7
1
q  tan (
)
7.36
X component Y component
3.89N
-6.06N
-5.19N
Rx=-7.36N
2.25N
-3.5N
3.0N
Ry=1.7N
Angle =-13o+180o = 167o
Concept Questions
• A stone is thrown horizontally from a cliff.
• How would the x distance change if the stone
was thrown twice as fast?
d x = vx t
dx = (2vx ) t
• dx distance would double
Concept Questions
• How would the v2y change if the stone was
thrown twice as fast?
• Vx and Vy are independent
• Since it was thrown horizontally, it would not
change.
Concept Questions
• How would the v2y change if the cliff was twice
as high?
vy 2  v
2
2
y1
 2gd
vy 2  2 gd
vy 2  2 g (2d )
v2 y  2 2 gd
The Partial Parabola
• Recall, this path has elevation and launch angle.
• The trajectory again has an apex.
• This is mathematically the most complex path.
Fire
Sample Partial Parabola Problem
• A cannon nestled in the side of a cliff (d1y = 20m)
fires a cannonball at 130m/s at a 40° angle.
– What are the components of the initial velocity?
Horizontal Component
v1x  v1 cos q
Fire
Vertical Component
v1 y  v1 sin q
v1x  130 ms  cos  40
v1y  130 ms  sin  40
v1x  99.6 ms
v1y  83.6 ms
v  v 2  2 gd
t
g
Sample Partial Parabola Problem
• A cannon nestled in the side of a cliff (d1y = 20m)
fires a cannonball at 130m/s at a 40° angle.
Fire
Partial Parabola
Variable
• You are trying to win a prize by throwing an
apple into a basket on top of a pedestal.
• The apple leaves your hand 1.00 m beneath
the top of the pedestal.
• The apple flies 3.09 m horizontally before
landing in the bottom of the basket.
• The apple’s maximum height was 1.26 m.
• What was the apple’s initial velocity
(magnitude and direction)?
a
v1
v2
d1
d2
t
x
y
Partial Parabola
• Find V1y knowing Dy max = 1.26m
2
2
2
v
vy 2  vy1  2gd
y 2  2 gd  v y1
vy 2  2gd  v
2
2
y1
vy1  (0 ms )2  2(9.8 sm2 )(1.26m)
vy1  4.97 ms
Dy (max) = 1.26m
Partial Parabola
Find the hang time knowing v1y and ending height d2y=1m
1 2
d 2 y  v1t  gt
2
v1 y  v1 y 2  4(.5)( g )(d )
2
b  b  4ac
t
x
2(.5)( g )
2a
1
a g
2
b  v1 y
c  d
t  .73s
Partial Parabola
Partial parabolas can be represented by two half parabolas
Total hang time is the time of the ½ parabola on the way up plus
the time of the ½ parabola on the way down.
t  .73s
t
v2 y  v1 y
g
t  .5s
t
2d y
g
t  .23s
The Partial Parabola
• If you look at this path carefully, you can see two
half parabolas, which simplifies things considerably.
• You still must consider the launch angle and the
components of the velocity when trying to solve.
q
Mortar Problems
• A mortar crew fires a projectile at
an enemy ammunitions storage
facility that is protected by a wall
located on top of a 200.0 m high
cliff.
• The ammunition is located a
horizontal distance of 314.68 m
from the mortar’s position.
• The projectile passes directly
over the wall at its maximum
height of 215.24 m.
• What was the projectile’s initial
velocity (magnitude and
direction)?
V
a
v1
v2
d1
d2
t
x
y