Question

A rocket is fired at a speed of 79.7 m/s from ground level, at an angle of 58.4^° above the horizontal. The rocket is fired toward an 10.5 m high wall, which is located 25.7 m away. By how much does the rocket clear the top of the wall?

Answer 1

Initial Velocity in x direction = 79.7 * cos(58.4) = 41.76 m/s
Initial Velocity in y direction = 79.7 * sin(58.4) = 67.88 m/s

Distance need to cover in horizontal direction = 25.7 m

Acceleration in horizontal direction = 0
Acceleration in Vertical direction = 9.8 m/s^2 downwards

Now time taken to cover horizontal distance t = 25.7/ 41.76 s = 0.62 s

Vertical Distance covered in time t

s = u*t - 0.5 * g * t^2
s = 67.88 * 0.62 - 4.9 * 0.62^2
s = 40.2 m

Rocket Clears the top wall by 40.2m - 10.5m
Rocket Clears the top wall by 29.7 m