well, you know that
y = -g/2v^2 sec^2 θ x^2 + tanθ x
so, plug in x=25 and subtract 21.4
A rocket is fired at a speed of 91.0 m/s from ground level, at an angle of 49.0 ° above the horizontal. The rocket is fired toward an 21.4-m high wall, which is located 25.0 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?
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