A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 59.7° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 28.5 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?

Horizontal problem:

u = 75 cos 59.7 = 37.8 m/s horizontal
t = 28.5/37.8 = .753 s

Vertical problem:
Vi = 75 sin 59.7 = 64.75 m/s initial
h = 64.75 t - 4.9 t^2
= 48.8 - 2.8 = 46 meters high
so
46 - 11 = 35 meters

To find out how much the rocket clears the top of the wall, we need to break down the rocket's motion into horizontal and vertical components.

First, we'll calculate the time it takes for the rocket to reach the wall. We can use the horizontal component of the rocket's initial velocity and the horizontal distance to find the time:

Horizontal component of initial velocity (Vx) = V * cos(θ)
Vx = 75.0 m/s * cos(59.7°) ≈ 37.89 m/s

Time taken to reach the wall (t) = distance / Vx
t = 28.5 m / 37.89 m/s ≈ 0.752 seconds

Next, we'll calculate the height of the rocket when it reaches the wall. We can use the vertical component of the rocket's initial velocity and the time taken to find the height:

Vertical component of initial velocity (Vy) = V * sin(θ)
Vy = 75.0 m/s * sin(59.7°) ≈ 64.66 m/s

Height when reaching the wall (h) = Vy * t + (0.5 * g * t^2)
Here, g represents the acceleration due to gravity, which is approximately 9.8 m/s^2.

h = (64.66 m/s * 0.752 s) + (0.5 * 9.8 m/s^2 * (0.752 s)^2)
h ≈ 48.53 m

Finally, to find out how much the rocket clears the top of the wall, we subtract the height of the wall from the height when reaching the wall:

Clearance = h - wall height
Clearance = 48.53 m - 11.0 m
Clearance ≈ 37.53 m

Therefore, the rocket clears the top of the wall by approximately 37.53 meters.