Arocket 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?

14.03 m

Explanation:

Given

initial velocity(u)=92 m/s

Launch angle

Height of wall=15 m

Located at x=27 m

maximum height

We know equation of trajectory of projectile is

for x=27 m

y=29.98-0.942=29.03 m

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

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.

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