PROJECTILE
MOTION
CHAPTER 3
PROJECTILE MOTION
An object launched into space without motive power of
its own is called a projectile.
If we neglect air resistance, the only force acting on a
projectile is its weight, which causes its path to deviate
from a straight line.
The projectile has a constant horizontal velocity and a
vertical velocity that changes uniformly under the
influence of the acceleration due to gravity.
HORIZONTAL PROJECTION
If an object is projected horizontally, its motion can
best be described by considering its horizontal and
vertical motion separately.
In the figure we can see that the vertical velocity and
position increase with time as those of a free-falling
body. Note that the horizontal distance increases linearly
with time, indicating a constant horizontal velocity.
3.19 A cannonball is projected horizontally with an initial velocity
of 120 m/s from the top of a cliff 250 m above a lake.
a. In what time will it strike the water at the foot of the cliff? UAM
v0x = 120 m/s
y = 250 m
v0y = 0
t
2y
2(250)

= 7.14 s
g
9.8
b. What is the x-distance (range) from the foot of the cliff to the
point of impact in the lake?
UM
x = vx t
= 120(7.14)
= 857 m
c. What are the horizontal and vertical components of its final
velocity?
vx = 120 m/s
vy = voy + gt
= 0 + 9.8 (7.14)
= 70 m/s
UM
UAM
d. What is the final velocity at the point of impact and its direction?
UAM
2
2
= 139 m/s

(
120
)

(
70
)
vR  v  v
2
x
70
  tan
120
1
2
y
= 30.2 below horizontal
v (139 m/s, 30.2)
3.20 A person standing on a cliff throws a stone with a horizontal
velocity of 15.0 m/s and the stone hits the ground 47 m from the
base of the cliff. How high is the cliff?
vx = 15 m/s
x = 47 m
vy = 0
UAM
x 47
t

v x 15
UM
= 3.13 s
y = ½ gt2
= ½ (9.8)(3.13)2
= 48 m
PROJECTILE MOTION AT AN ANGLE
The more general case of projectile motion occurs when
the projectile is fired at an angle.
Problem Solution Strategy:
1. Upward direction is positive. Acceleration due to gravity (g) is downward thus
g = - 9.8 m/s2
2. Resolve the initial velocity vo into its x and y components:
vox = vo cos θ
voy = vo sin θ
3. The horizontal and vertical components of its position at any instant is given
by:
x = voxt
y = voy t +½gt2
4. The horizontal and vertical components of its velocity at any instant are given
by: vx = vox
vy = voy + gt
5. The final position and velocity can then be obtained from their components.
3.23 An artillery shell is fired with an initial velocity of 100 m/s at an
angle of 30 above the horizontal. Find:
UAM
a. Its position and velocity after 8 s
UM
vo = 100 m/s, 30
t=8s
g = - 9.8 m/s2
x = vox t
= 86.6(8)
= 692.8 m
y = voy t + ½ gt2
= 50(8) + ½ (-9.8)(8)2
= 86.4 m
vox = 100 cos 30
= 86.6 m/s
voy = 100 sin 30
= 50 m/s
vx = vox = 86.6 m/s
vy = voy + gt
= 50 + (-9.8)(8)
= - 28.4 m/s
b. The time required to reach its maximum height
At the top of the path:
vy = 0
vy = voy + gt
t
c. The horizontal distance (range)
Total time
T = 2t
= 2(5.1)
= 10.2 s
 voy
g
UAM
 50

= 5.1 s
 9 .8
UM
x = vox t
= 86.6(10.2)
= 883.7 m
3.24 A baseball is thrown with an initial velocity of 120 m/s at an
angle of 40above the horizontal. How far from the throwing point
will the baseball attain its original level?
vo = 120 m/s, 40
g = - 9.8 m/s2
At top vy = 0
UAM
t
 voy
g
vox = 120 cos 40
= 91.9 m/s
voy = 120 sin 40
= 77.1 m/s
 77.1

= 7.9 s
 9 .8
UM
x = vox (2t)
= 91.9(2)(7.9)
= 1452 m
3.25 A plastic ball that is released with a velocity of 15 m/s stays in
the air for 2.0 s.
a. At what angle with respect to the horizontal was it released?
vo = 15 m/s
t=2s
UAM
time to maximum height = 1 s
at the top vy = 0
vy = voy + gt
vo sin 
t
g
tg
sin  
vo
  sin
1
9.8(1)
= 40.8º
15
b. What was the maximum height achieved by the ball?
y = voy t +½ gt2
= (15)(sin 40.8º)(1) + ½ (-9.8)(1)2
= 4.9 m
UAM
3.26 Find the range of a gun which fires a shell with muzzle
K2D
velocity vo at an angle θ .
At top vy = 0
vy = voy + gt
= vo sin θ - gt
vo sin 
t
g
Total time = 2t
x = vxt  v cos   2vo sin  
o



2
o
g
2v
x
(sin  cos  )
g

sin θ cos θ = ½ sin 2θ
2
o
2v
x
(sin  cos  )
g
2
o
v
x  sin 2
g
b. What is the angle at which the maximum range is possible?
Maximum range is 45 since 2θ = 90
sin θ cos θ = ½ sin 2θ
2
o
2v
x
(sin  cos  )
g
2
o
v
x  sin 2
g
b. What is the angle at which the maximum range is possible?
Maximum range is 45 since 2θ = 90
c. Find the angle of elevation  of a gun that fires a shell with
muzzle velocity of 120 m/s and hits a target on the same level
but 1300 m distant.
vo = 120 m/s
x = 1300 m
2
o
v
x  sin 2
g
gx 9.8(1300)
sin 2  2 
= 0.885
2
vo
(120)
sin-1(2θ)= 62
θ = 31
PROJECTILE MOTION
An object launched into space without motive power of
its own is called a projectile.
If we neglect air resistance, the only force acting on a
projectile is its weight, which causes its path to deviate
from a straight line.
c. Find the angle of elevation  of a gun that fires a shell with
muzzle velocity of 120 m/s and hits a target on the same level
but 1300 m distant.
vo = 120 m/s
x = 1300 m
2
o
v
x  sin 2
g
gx 9.8(1300)
sin 2  2 
= 0.885
2
vo
(120)
sin-1(2θ)= 62
θ = 31
c. Find the angle of elevation  of a gun that fires a shell with
muzzle velocity of 120 m/s and hits a target on the same level
but 1300 m distant.
vo = 120 m/s
x = 1300 m
2
o
v
x  sin 2
g
gx 9.8(1300)
sin 2  2 
= 0.885
2
vo
(120)
sin-1(2θ)= 62
θ = 31
PROJECTILE MOTION
An object launched into space without motive power of
its own is called a projectile.
If we neglect air resistance, the only force acting on a
projectile is its weight, which causes its path to deviate
from a straight line.
c. Find the angle of elevation  of a gun that fires a shell with
muzzle velocity of 120 m/s and hits a target on the same level
but 1300 m distant.
vo = 120 m/s
x = 1300 m
2
o
v
x  sin 2
g
gx 9.8(1300)
sin 2  2 
= 0.885
2
vo
(120)
sin-1(2θ)= 62
θ = 31