Question

A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 60.0° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 32.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?

Answer 1

Given values:

Vo = 75.0 m/s
θ = 60.0°
g = 9.81 m/s^2
Y = 11.0 m
r = 32.0 m

Find initial horizontal and vertical components of velocity

Vox = Vo * cosθ
Vox = (75.0 m/s) * cos(60.0)
Vox = (75.0 m/s) * (0.5)
Vox = 37.5 m/s

Voy = Vo * sinθ
Voy = (75.0 m/s) * sin(60)
Voy = (75.0 m/s) * (0.866)
Voy = 64.95 m/s

Time it takes to reach the wall

t = r / Vox
t = (11.0 m) / (37.5 m/s)
t = 0.293 s

Total height at 0.293seconds (11.0 m downrange and over the wall)

H = [ Voy * t ] - [ 0.5 * g * t^2 ]
H = [ (64.95 m/s) * (0.293 s) ] - [ 0.5 * (9.81 m/s^2) * (0.293 s)^2 ]
H = [ 19.03 m ] - [0.421 ]
H = [ 19.03 m ] - [ 0.421 m ]
H = 18.609 m

So it clears it by

h = H - Y
h = (18.609 m) - (11.0 m)
h = 7.609 m