A rocket is fired at a speed of 70.0 m/s from ground level, at an angle of 40.0 ° above the horizontal. The rocket is fired toward an 14.0-m high wall, which is located 21.0 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?

To solve this problem, we can break down the motion of the rocket into horizontal and vertical components. Let's first calculate the time it takes for the rocket to reach the wall, and then calculate its vertical displacement.

1. Calculate the time of flight:
To find the time it takes for the rocket to reach the wall, we need to use the horizontal component of its initial velocity.

Horizontal component of velocity (Vx) = velocity * cos(angle)
Vx = 70.0 m/s * cos(40.0°)

Now, we can calculate the time it takes for the rocket to reach the wall using the equation:
Time = distance / Vx
Time = 21.0 m / Vx

2. Calculate the vertical displacement:
To find the vertical displacement, we need to consider the vertical component of the rocket's initial velocity and the acceleration due to gravity.

Vertical component of velocity (Vy) = velocity * sin(angle)
Vy = 70.0 m/s * sin(40.0°)

Using the equation of motion in vertical direction:
Vertical displacement (d) = Vy * time + (1/2) * acceleration due to gravity * time^2
d = (Vy * time) + (0.5 * (-9.81 m/s^2) * time^2)

Finally, to find how much the rocket clears the top of the wall, we subtract the height of the wall from the vertical displacement:
Clearance = d - 14.0 m

Now, let's plug in the values and calculate:

Vertical component of velocity (Vy) = 70.0 m/s * sin(40.0°)
Vx = 70.0 m/s * cos(40.0°)
Time = 21.0 m / Vx
d = (Vy * time) + (0.5 * (-9.81 m/s^2) * time^2)
Clearance = d - 14.0 m

Evaluate the above equations to find the clearance of the rocket above the top of the wall.