A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 60.5° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 22.5 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?
V0x= V0Cos (angle)
V0y= V0Sin (angle)
The Attempt at a Solution
To solve this problem I first used the first two equations to find the V0x and V0y. I ended up with 36.93 for x and 65.276 for y.
From there I used the 3rd equation considering that acceleration (a)=0 so I ended up with x=V0t or 22.5=(35.93)t. I ended up with .609 seconds.
From here I used the last equation (considering that acceleration is constant) and found that Y(t)=V0y(t) or y(t)=65.276 x .609.
My answer was 39.75. I subtracted 11 from that and ended with 28.75 which is not the correct answer.