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?

Question

bobpursley · Answer

Horizontal distance=horzvelocity*time time= distance/velocity= 25/(75cos64.7) figure time, then put it here. Vertical: h=75*sin64.7t-4.9t^2 how much higher is h than 11 m.

hy · Answer

thanks

liz · Answer

where does the 4.9 come from?

tom · Answer

h=49.62m 49.62m-11m for the wall clear of the top = 38.62 4.9 its from multiplying 1/2(-9.8m/s^2) 1/2(ay)