physics
posted by Cody .
A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 68.6° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 26.0 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?

The height of the rocket fired at speed v at angle a is given by
y(x) = g*sec^2(a)/2v^2 x^2 + x*tan(a)
= .0065x^2 + 2.55x
y(26) = 61.9m so it clears the 11m wall by 50.9m
Makes sense, since if the rocket went in a straight line at an angle of 68.6°, at 26m out it would be 66.3m high.
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