PROJECTILE MOTION
Chapter 3, sections 3.1-3.3, 3.5
EXAMPLES OF PROJECTILE MOTION
PROJECTILE MOTION
vx
v
vx: Horizontal Velocity.
Once in air, vx does not
change since there is
no force pushing or
pulling horizontally.
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.
v
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 yt  1 / 2 a yt
 1 / 2a yt
t
2
2
2y
ay
Time it takes to hit the
ground is completely
unaffected by the
horizontal velocity
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  v0 xt
Vertical (FREE FALL)
y  v yt
v y  v0 y  a yt
vy 
1
2
(v0 y  v y )
 y  v0 yt 
v v
2
y
2
0y
1
2
a yt
2
 2a yy
Remember…
To work projectile problems…
• …resolve the initial, launch velocity into
horizontal and vertical components.
Vo

V0x = Vo cos 
V0y = Vo sin 
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
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
vfy = -v0y at landing
Initial Horizontal velocity
v 0 x  v x  v 0 cos 
Initial Vertical velocity
(determines air time)
v 0 y  v 0 sin 
x
vf
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)
y
x
v x  v0
v0 y  0
 x  67 . 1 m
vy
t
a y   9 .8 m / s
 y   12 . 5 m
12.5 m
t = 1.6 s
t
v0 = 41.9 m/s
vf = 44.7 m/s, 20.50 below the
horizontal
67.1 m
2
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
v 0  40 m / s
60
0
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  90 m v y
2
t
a y   9 .8 m / s
 y B   50 . 0 m
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
v0 y  0
v x  28 . 5 m / s
vy
 x  90 m
2
t
a y   9 .8 m / s
 y B   50 . 0 m
t
50.0 m
vf = ?
90.0 m
EXAMPLE
What horizontal firing (muzzle) velocity
splashes the cannonball into the pond?
x - component
v x  v0
 x  95 m
t
v0
y - component
v0 y  0
vy
ay
40 m
t = 2.86 s
v0 = 33.2 m/s
95 m
 y   40 m
t
EXAMPLE
Where does the ball land? How far has the
truck traveled in that time?
Ball
x
v x  30 m / s
x
t
v = 10 m/s
y
v 0 y  10 m / s
vy
a y   9 .8 m / s
2
y  0
v = 30 m/s
t
t = 2.0 s
x = 60 m
B1
B2
http://zamestnanci.fai.utb.cz/~schauer/PhET1.0/simulations/simsa79a.html?sim=Projectile
_Motion
Projectile Launched from Any Angle
Horizontal velocity
v 0 x  v 0 cos 
Vertical velocity
v 0 y  v 0 sin 
When starting and landing heights are the same:
Time of flight
Range
t 
 2v0 y

 2 v 0 sin 
ay
ay
 2 v 0 cos  sin 
2
x 
ay
 v 0 sin
2
Max height
 y max 
2a y
 v 0 sin 2
2

ay
2

Max at 450
All depend on angle
and initial velocity
Remember…
To work projectile problems…
• …resolve the initial velocity into components.
Vo

Vo,x = Vo cos 
Vo,y = Vo sin 
DO NOW
Sketch the following motion graphs for a horizontally launched
projectile.
x
y
t
t
vx
vy
t
t
ax
ay
t
t
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
v 0  6 .6 m / s

 x  2 .0 m
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
v x  v 0 cos 
 x  25 . 8 m
t  1 . 98 s
y
vx = 13.0 m/s
v0y = 11.13 m/s
v0 = 17.1 m/s, 40o
v 0 y  v 0 sin 
vy
a y   9 .8 m / s
2
 y  2 . 44 m
v0  ?
t  1 . 98 s
2.44 m

25.8 m
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  90 m v y
2
t
a y   9 .8 m / s
 y B   50 . 0 m
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
v0 y  0
v x  28 . 5 m / s
vy
 x  90 m
2
t
a y   9 .8 m / s
 y B   50 . 0 m
t
50.0 m
vf = ?
90.0 m
EXAMPLE
A player kicks a football from ground level with an initial velocity
of 27.0 m/s 300 above the horizontal. Assume air resistance is
negligible.
a) Find the ball’s hang time
b) Find the ball’s maximum height
c) Find the ball’s range
d) Find the impact velocity
Player then kicks the ball with the same speed but at a 60 degree
angle. Find all the above
v 0  27 . 0
30
0
300
a) t = 2.76 s
b) ymax = 9.30m
c) range=64.5m
d) vf=27m/s -300
600
a) t = 4.78 s
b) ymax = 27.9m
c) range=64.5m
d) vf=27m/s -600
EXAMPLE
Agent Tim flying a constant 185 km/hr horizontally in a low
flying helicopter, wants to drop a small explosive onto a master
criminal’s automobile, traveling 145 km/hr on a level highway
88.0 m below. At what angle (with the horizontal) should the car
be in his sights when the bomb is released? 62.1o
185 km/h

88.0 m
145 km/h
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
“Projectile motion and parabolas”: Football kick, projectile motion
http://www.nbclearn.com/portal/site/learn/nfl
/cuecard/50974/
Wile E. Coyote Physics Problem
The determined Wile E. Coyote is out once more to try to
capture the elusive roadrunner. The coyote wears a new pair of
Acme power roller skates, which provide a constant horizontal
acceleration of 15 m/s^2. The coyote starts off at rest 70m from
the edge of a cliff at the instant the roadrunner zips by in the
direction of the cliff. A) If the roadrunner moves with constant
speed, find the minimum speed the roadrunner must have in
order to reach the cliff before the coyote. ____m/s B) If the cliff is
100m above the base of a canyon, find where the coyote lands in
the canyon. (Assume that his skates are still in operation when
he is in "flight" and that his horizontal component of acceleration
remains constant at 15m/s^2. ____m from the base I don't need
answers, but I do need help on how to get there.