Cecilia is taking Physics. I am an engineer. She won’t ask me for help. She would rather post the question on her blog. I’m giving it to her anyway.

Here’s the question:

A rocket is fired at a speed of 94.0 m/s from ground level, at an angle of 43.0 degrees above the horizontal. The rocket is fired toward an 31.8-m high wall, which is located 40.0 m away. By how much does the rocket clear the top of the wall?

Here’s the answer:

This question is bogus – no rocket achieves constant velocity at t=0 seconds or even before d=40.0 meters. It experiences constant acceleration, which produces an exponential curve when graphed over time. The author of this “question” ought to be strapped to a Shuttle Booster Rocket to let him see what real life is actually like. However, for the purposes of pleasing academia . . .

Assuming a constant velocity of 94.0 meters/second

Actual height when rocket reaches d=40.0 meters is found using the tangent function:

tan 43 degrees = 0.9325 = y(height) / x (distance) = y(height) / 40.0

Solving for y:

y = 0.9325*40.0 = 37.3 meters

Minimum y (height) required to clear wall > 31.8 meters

Therefore, the make-believe rocket clears the wall by 5.4 meters. That's assuming said rocket has no diameter and is an infinitely small line in space, as Mr. Author would have us assume also. If the rocket was, in fact, as large as a Shuttle Booster, it would only clear the wall by 1.7 meters.

See Dad, I really did learn something in college.

*although my physics, geometry, and trigonometry are intact, apparently my basic math skills have degenerated to the point of retardation. The answer should be 5.50 meters, not 5.40. See this for the explanation