A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 68.0° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 26.5 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?

Please show some evidence of your own work.

To solve this problem, we can break it into horizontal and vertical components. We need to find the vertical distance the rocket clears the top of the wall.

First, let's find the time it takes for the rocket to reach its maximum height. We can use the vertical component of the initial velocity and acceleration due to gravity.

Vertical component of initial velocity (Viy) = V * sin(θ)
= 75.0 m/s * sin(68.0°)
≈ 68.755 m/s

Acceleration due to gravity (g) = 9.8 m/s^2

Using the kinematic equation:
Vfy = Viy + (-g)t

At maximum height, Vfy = 0 m/s (because the rocket momentarily stops at its highest point). Therefore, we can solve for t:

0 = 68.755 m/s - 9.8 m/s^2 * t

Solving this equation, we get:
t ≈ 7.02 seconds

Now, let's find the vertical distance the rocket travels during this time. We can use the equation:

Δy = Viy * t + (1/2) * (-g) * t^2

Plugging in the values:
Δy = 68.755 m/s * 7.02 s + (1/2) * (-9.8 m/s^2) * (7.02 s)^2
≈ 241.88 m

So, the rocket clears the top of the wall by approximately 241.88 meters.