1. The problem statement, all variables and given/known data 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? 2. Relevant equations V0x= V0Cos (angle) V0y= V0Sin (angle) x=1/2at(squared)+V0t X(t)=vt 3. 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. Thanks.