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
Vertical
vHi = vi cos
vvi = vi sin
vHi = vHf = vH
vvf = vVi + g t
dH = vH t
a = 0 m/s2
dv = vvi t + 1/2 g t2
a = g = -9.8 m/s2
Example Problem
A projectile is launched from a height of 44.1 m with a initial
horizontal speed of 20 m/s. a) How long is it the air? b) how far
does it travel horizontally before it hits the ground?
vvi = 0 m/s
vH = 20 m/s
dv = - 44.1 m g = - 9.8 m/s2
a) dv = vvi t + 1/2 g t2
20m/s
- 44.1 m = (0 m/s) t + 1/2 (-9.8 m/s2) t2
t =  (2 (-44.1 m) / -9.8
m/s2)
b) dH = vH t
dH = (20 m/s) (3s) = 60m
= 3s
- 44.1 m
Example Continued
c) What velocity does the projectile hit the ground?
vvi = 0 m/s
vH = 20 m/s

vvf
vH
v
dv = -44.1 m g = - 9.8 m/s2 t = 3s
vVf = vVi + g t
vVf = (0 m/s) + (-9.8 m/s2) (3s)
= - 29.4 m/s
v =  (vvf2 + vH2) = (- 29.4 m/s)2 + (20m/s)2 = 35.6 m/s
 = tan-1 (vvf / vH) below the horizontal
 = tan-1 (29.4 m/s / 20 m/s) = 55.8 0 below the horizontal
v = 36 m/s @ 560 below the horizontal
Example Continued
or… can solve for speed
using conservation of energy

vvf
vH
vBottom
vTop= 20 m/s hTop = 44.1 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 (20 m/s)2 + (-(-9.8 m/s2) (44.1m) = 1/2 vBottom2 + (-(-9.8m/s2)(0m)
632.2 m2/s2 = 1/2 vBottom2
vBottom =  2(632.2 m2/s2) = 35.6 m/s