the height h is
h(x) = x tanθ - g/(2(v cosθ)^2) x^2
= .7x - .0008 x^2
h(32) = 21.58
So, it clears the wall by 21.6-15.2 = 6.4m
A rocket is fired at a speed of 96.0 m/s from ground level, at an angle of 35.0 ° above the horizontal. The rocket is fired toward an 15.2-m high wall, which is located 32.0 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?
2 answers
940