Projectile Motion - Horizontal launches
The scale on the left is 1cm = 5m
Draw the positions of the dropped ball every
second that it is in the air. Neglect air
resistance and assume g = 10 m/s2.
Estimate the number of seconds that the
ball is in the air
Four positions of a ball thrown horizontally
without gravity are shown. Draw the
positions of the ball with gravity. Describe
the path taken by the ball.
It is parabolic
How is motion in the vertical direction
affected by motion in the horizontal
direction?
They are independent of each other
Projectile Motion
Rules:
1. Vertical Motion is effected by gravity
2. Horizontal motion is constant
3. Motion in the horizontal direction is
independent of the motion in the vertical
direction.
4. Assume no air resistance
Physical Expressions
Horizontal
vHi = vHf = vH
dH = vH t
a = 0 m/s2
Vertical
vvf = vVi + g t
dv = vvi t + 1/2 g t2
a = g = -9.8 m/s2
Example A Falling Care Package
The airplane is moving horizontally with a constant velocity of
+115 m/s at an altitude of 1050m. Determine the time required
for the care package to hit the ground.
Example A Falling Care Package
dv
av
-1050 m
- 9.8 m/s2
vvf
vvi
0 m/s
tair
?
Example A Falling Care Package
dv
g
-1050 m
-9.80 m/s2
vvf
dv  vvi t  gt
1
2
2dv
t 

g
2
vvi
tair
0 m/s
?
dv  gt
1
2
2  1050 m 
9.80 m s
2
2
 14.6 s
Example A Falling Care Package
What is the horizontal distance traveled by the care package?
dH
Example A Falling Care Package
dH
dH
?
aH
0 m/s2
vH
tair
115 m/s
14.6 s
Example A Falling Care Package
dH
aH
vH
tair
?
0 m/s2
115 m/s
14.6 s
dH  vH tair  115m s 14.6 s 
 1679 m
Example Problem
Supposed that you are an accident investigator and are asked to figure out whether the car was
speeding before it crashed through the rail of the bridge into the mudbank as shown below. The
speed limit on the bridge is 55 mph (24m/s). What’s your conclusion?
?
vvi = 0 m/s dH = 24m dv = - 4.9 m g = - 9.8 m/s2 vH = ?
dv = vvi t + 1/2 g t2
- 4.9 m = (0 m/s) t + 1/2 (-9.8 m/s2) t2
- 4.9 m
t =  (2 (-4.9 m) / -9.8 m/s2) = 1 s
dH = vH t
24m = vH (1s)
24 m
So… vH = 24m / 1s = 24 m/s
The car loses speed when it went through the railing so it must have been speeding
Example Continued

What speed did the car hit the mudbank?
Easiest to solve using conservation of energy
vTop= 24 m/s hTop = 4.9 m hBottom = 0m g = -9.8 m/s2 vBottom = ?
KETop + GPETop = KEBottom + GPEBottom
1/2 m vTop2 + (-mghTop) = 1/2 m vBottom2 + (-mghBottom)
1/2 vTop2 + (-ghTop) = 1/2 vBottom2 + (-ghBottom)
1/2 (24 m/s)2 + (-(-9.8 m/s2) (4.9m) = 1/2 vBottom2 + (-(-9.8m/s2)(0m)
336 m = 1/2 vBottom2
vBottom =  2(336 m) = 26 m/s
vvf
vH
vBottom