physics
posted by hy .
A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 64.7¡ã above the horizontal. The rocket is fired toward an 11.0 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?

Horizontal
distance=horzvelocity*time
time= distance/velocity= 25/(75cos64.7)
figure time, then put it here.
Vertical:
h=75*sin64.7t4.9t^2
how much higher is h than 11 m. 
thanks

where does the 4.9 come from?

h=49.62m
49.62m11m for the wall
clear of the top = 38.62
4.9 its from multiplying 1/2(9.8m/s^2)
1/2(ay)