PROJECTILE MOTION
Chapter 3, sections 3.1-3.3, 3.5
EXAMPLES OF PROJECTILE MOTION
PROJECTILE MOTION
vx
v1
vx: Horizontal Velocity.
Once in air, vx does not
change since there is
no force pushing or
pulling horizontally.
v2
vy: Vertical Velocity. Once
in air, vy changes because
of the force of gravity. The
projectile accelerates at the
rate of 9.8 m/s2 downward.
v3
PM simulation
PROJECTILE MOTION
• A projectile is an object shot through the air upon which the
only force acting is gravity (neglecting air resistance).
• The trajectory (path through space) or a projectile is parabolic.
Time of flight
(for horizontally launched
projectile):
y  v0 y t  1 / 2a y t 2
 1 / 2a y t 2
2y
t
ay
Time it takes to hit the
ground is completely
unaffected by the
horizontal velocity,
only depends on height
The horizontal part of a projectile motion problem
is a constant velocity problem.
•
• The vertical part of a projectile motion problem is a
free fall problem.
• Horizontal and vertical parts of the motion are
independent
• To solve a projectile motion problem, break up the
motion into x and y components.
Kinematics Equations
for uniformly accelerated motion
Horizontal (ax=0)
 
x  v x t
Vertical (FREE FALL)

y  vav y t
 

v y  v0 y  a y t



1
vav y  2 (v0 y  v y )
 
 2
1
y  v0 y t  2 a y t
2 2
 
v y  v0 y  2a y y
Some Common Projectile Motion
Problems
Horizontal Launch
v0
When object is
launched horizontally:
v0y = 0
x, Range
Angle Launch on Level Ground
When object returns to
the same height from
which it was launched:
v0
x, Range
y = 0
Remember…
To work projectile problems…
• …resolve the initial, launch velocity into
horizontal and vertical components.
Vo

vx = vo cos 
v0y = vo sin 
Projectile Launched from Any Angle
y
vt
vty
v0
v0y
vx
vx

vx
vx
ymax = height
a = constant
along y
At max height, vy = 0
vx
vx
vx
v = constant along x
xmax = range
Initial Horizontal velocity
v0 x  vx  v0 cos 
x
vf
vy = -v0y at same
height landing
Initial Vertical velocity
(determines air time)
v0 y  v0 sin 
PM simulation
EXAMPLE
A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which
is located 67.1 m from the base of the bluff. She launches a horizontal
shot that lands in the hole on the fly. The gallery erupts in cheers.
a) How long was the ball in the air?
b) What was the ball’s initial velocity?
c) What was the ball’s impact velocity (velocity right before landing)
vx  v0
x
x  67.1m
t
12.5 m
y
v0 y  0
vy
a y  9.8m / s 2
y  12.5m
t
t = 1.6 s
v0 = 41.9 m/s
v = 44.7 m/s, 210 below the
horizontal
67.1 m
EXAMPLE
In the circus, a clown is launched from a cannon at
40 m/s, 60o from the horizontal. Where should the
other clowns hold the net so that the projectile
clown lands unharmed (at the same level)?
t = 7.07 s
x = 141 m
60
0
vx
v0y
EXAMPLE
Where does the ball land? How far has the
truck traveled in that time?
Ball
x
vx  30m / s
x
t
v = 10 m/s
y
v0 y  10m / s
v fy
a y  9.8m / s 2
y  0
t
v = 30 m/s
t = 2.0 s
x = 60 m
B1
B2
EXAMPLE
If Agent Tim is flying at a constant velocity of 86.0 m/s horizontally in a
low flying helicopter at an altitude of 125 m. His mission is to drop a 50 kg
explosive onto a master criminal’s car. How far before the car should he
release the 50 kg bomb so that it lands on the car? (ignore air resistance)
86 m/s
125 m
x = ?
Bomb
v x  86.0m / s v0 y  0
x 
vy
t
a y  10m / s 2
y  125m
t
430m
Cargo Drop
EXAMPLE A cannon ball is fired out of a cannon at
muzzle velocity of 30 m/s, 40o above the horizontal. The
cannonball is fired from a 100.0 m high cliff. How far
from the base of the cliff will the cannonball land? What
is its final velocity, right before landing?
y
x
40o
v x  22.3m / s
v y = -48.32m/s
x  ?
t
v0 y  19.3m / s
= 6.90s
a y  9.8m / s 2
y  100.0m
100.0
100.0mm
t = 6.90s
x = 153.9 m
vx
x
vy
v = 53.2m/s,
65o below the
horizontal
EXAMPLE A professional soccer player is 25.8 m from
the goal and kicks a hard shot at ground level. The ball
hits the cross bar on its way down, 2.44 m above the
ground, 1.98 s after being kicked. What was the initial
velocity of the ball (find the x- and y- components)?
x
vx  v0 cos 
x  25.8m
t  1.98s
v0  ?
y
v0 y  v0 sin 
vy
vx = 13.0 m/s
v0y = 11.13 m/s
v0 = 17.1 m/s, 40o
a y  9.8m / s 2
y  2.44m
t  1.98s

2.44 m
25.8 m
EXAMPLE
What horizontal firing (muzzle) velocity
splashes the cannonball into the pond?
x-dir
v0
y-dir
v x  v0
v0 y  0
x  95m
vy
t
a y  10m / s 2
y  40m
40 m
t = 2.86 s
v0 = 33.2 m/s
95 m
t
Great simulations of Projectile Motion
Any type of projectile (by Mr Walsh) http://tube.geogebra.org/student/m167259
https://phet.colorado.edu/sims/projectileProjectile over level ground
motion/projectile-motion_en.html
Exploring Concepts: Several projectiles with
same range OR
Same max height OR
Same initial speed
http://tube.geogebra.org/student/m224687
For the case of projectiles launched on level ground, you found that
the initial conditions of launch speed, v0, and launch angle, ,
affect the projectile’s flight time (t), maximum height (ymax), and horizontal range,
(x) in the following ways.
1. Time of flight increases with v0y, the vertical component of v0
2. Max height increases with v0y, vertical component of v0
3. Horizontal Range affected by both components of v0 so that if v0 stays the same,
max range is at 45o and
2 complimentary launch angles at the same speed gave the same range
For a projectile launched over level ground
DERIVE EXPRESSIONS for
time of flight (t),
max height (ymax), and
horizontal range (x)
in terms of v0, , and g.
Confirm that the derived equations
predict the observations above.
For the range equation Use the
trig identity : 2sincos = sin(2)
All launched at
same speed, v0
Projectile Launched from
Any Angle Over Level
Ground
Shown are all
launched at
same speed, v0
simulation
exploring concepts simulation
When starting and landing heights are the same:
Time of flight
Max height
Range
t
 2v0 y
ymax
ay
2v0 sin 

g
v02 sin 2 

2g
All depend on angle
and initial velocity
 2v02 cos  sin  v02 sin 2
x 

ay
g
Max at 450
Projectile Launched from Any Angle Over Level Ground
sin vs 
t
 2v0 y
ay

2v0 sin 
g
1.2
1
sin
Time of flight
Max height
ymax
0.8
0.6
0.4
0.2
v02 sin 2 

2g
0
0
15
30
45
60
 (degrees)
75
90
sin(2) vs 
1.2
v sin 2
x 
g
1
sin(2)
Range
2
0
0.8
0.6
0.4
0.2
0
0
15
30
45
60
 (degrees)
75
90
Rank time in air from greatest to least:
Red = green = blue
If the time in air for all was 6 sec
a) What is the initial vertical velocity 30 m/s
b) What is the max height? 45m (vavt)
For a Horizontally Launched Projectile
a) Derive an expression for the flight time, t, and horizontal range,
x, of a horizontally launched projectile in terms of the initial
conditions: v0, h and g.
b) If the projectile was launched horizontally from double the
height, what would happen to the flight time and range?
v0
y
x
Time of Flight
2h
t
g
Horizontal (x)
Vertical (y)
v x  v0
v0 y  0
x
vy
t
a y  g
y  h
Range
x  v0t  v0
t
2h
g
DO NOW
Sketch the following motion graphs for a horizontally launched
projectile.
x
y
t
t
vx
vy
t
t
ax
ay
t
t
Rank flight time from greatest to least:
m
m
B=E > A=D=F > C
v0
v0
m ½v
0
2h
h
(B)
(A)
2m
2m
h
v0
(C)
2v0
m
2h
(D)
½h
(E)
h
2v0
(F)
Rank range from greatest to least:
E > B> A > D > C
m
v0
m
2v0
m ½v
0
h
h
(A)
2m
2m
h
½v0
(B)
2v0
2h
(D)
(E)
½h
(C)
DO NOW
A fire hose held near the ground shoots water at a
speed of 6.6 m/s. At what angle(s) should the nozzle
point in order that the water land 2.0 m away? Why
are there 2 different angles?
 = 13.70, 76.30
v0  6.6m / s

x  2.0m
EXAMPLE
A player kicks a football from ground level with an initial velocity of 27 m/s
at an angle of 300 above the horizontal.
a) Determine the ball’s hang time (total time in air)
b) Determine the ball’s maximum height
c) Determine the ball’s range
d) What is the ball’s impact velocity
e) The player then kicks the ball with the same speed but at a 60 degree
angle. Which quantities change, which stay the same?
v0  27.0
300
300
a) t = 2.76 s
b) ymax = 9.30m
c) range=64.5m
d) v=27m/s -300
600
a) t = 4.78 s
b) ymax = 27.9m
c) range=64.5m
d) v=27m/s -600
EXAMPLE- Projectile over Level ground
A player kicks a football from ground level with an initial velocity of v0 at
an angle of  above the horizontal. The ball lands at the same height.
Neglect air resistance.
a) Derive an expression for the ball’s hang time (in terms of v0, , and g)
b) Derive an expression for the ball’s maximum height (in terms of v0, , and g)
c) Derive an expression for the ball’s range (in terms of v0, , and g)
Use the trig identity 2sincos = sin(2)
d) Find an expression for the impact velocity
e) If the launch velocity of the football is 27 m/s, 300 above the horizontal,
determine the ball’s hang time, max height, range and impact velocity.
f) Player then kicks the ball with the same speed but at a 60o angle.
Which quantities change, which stay the same?
2v0 sin 
t
g
x 
300
ymax 
2v 20 cos  sin 

v 20 sin 2 
2g
v 20 sin( 2 )
g
g

v  v0 , o below the horizontal
EXAMPLE A movie stunt driver on a motorcycle
speeds horizontally off a 50.0 m high cliff. How fast
must the motorcycle leave the cliff top if it is to land on
level ground below, 90.0 m from the base of the cliff
where the cameras are? What is its final velocity, right
before landing?
vo = 28.5 m/s
v0 = ?
50.0 m
x
y
v0 y  0
vx  ?
x  90m v fy
t
a y  9.8m / s 2
y B  50.0m
t
90.0 m
EXAMPLE
What is its final velocity, right before landing?
vf = 42.3 m/s, -47.70 from x-axis
x
y
v x  28.5m / s v0 y  0
x  90m
v fy
t
a y  9.8m / s 2
y B  50.0m
t
50.0 m
vf = ?
90.0 m
DO NOW
Frustrated with the book you are reading, you open the second
story classroom window and violently hurl your book out the
window with a velocity of 18 m/s at an angle of 35o above the
horizontal. If the launch point is 6 m above the ground,
a) how far from the building will the book hit the parking lot?
b) What is the final velocity of the book right before it hits the
ground?
x = 37.9 m
vf = 21 m/s, 45.40 below the x-axis