Question

A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 60.0° above the horizontal. The rocket is fired toward an 11-0m-high wall, which is located 27.0 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?

Answer 1

Answer:

ΔH = 33.17m

Explanation:

By knowing the amount of time it takes the rocket to travel the horizontal 27m, we will be able to calculate the height when x=27m. So:

$X = V_{o}*cos(60)*t$   where $X=27m; V_{o}=75m/s$

Solving for t:

t=0.72s

Now we calculate the height of the rocket:

$Y_{f}=V_{o}*sin(60)*t-\frac{g*t^{2}}{2} = 44.17m$

If the wall was 11m-high, the difference is:

ΔH = 33.17m