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 11-m 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.
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?
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