Question

A rocket is fired at a speed of 92.0 m/s from ground level, at an angle of 48.0 ° above the horizontal. The rocket is fired toward an 15.0-m 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:14.03 m

Step-by-step explanation:

Given

initial velocity(u)=92 m/s

Launch angle
`=48^(\circ)`

Height of wall=15 m

Located at x=27 m

maximum height
`h=(u^2sin^\theta )/(2g)=(92^2(sin48)^2)/(2* 9.8)`

`h_(max)=238.48`

We know equation of trajectory of projectile is

`Y=x\tan\theta -(gx^2)/(2u^2(cos\theta )^2)`

for x=27 m

`y=27\tan48-(9.8* 27^2)/(2* 92^2* (cos48)^2)`

y=29.98-0.942=29.03 m

Thus it passes at a height of 29.03-15 =14.03 m above wall