Math/ Science?

Question

A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 64.3° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 30.5 m away. By how much does the rocket clear the top of the wall?

Answer 1

v(zero horz) = 75cos64.3 = 32.542m/s
v(zero vert) = 75sin64.3 = 67.581m/s
Horizontal:
D = rt
t = D/r = 30.5/32.542 = 0.9378sec
Vertical:
s = s(zero) + v(zero)t + (1/2)gt^2
s = (67.581)(0.938) -(4.9)(0.938)^2
s = 59.079
Clears the top of the wall by:
59.080 - 11.0 = 48.080m