A rocket is fired at a speed of 75.0 m/s from ground level, at anangle of 66.6°

Question

A rocket is fired at a speed of 75.0 m/s from ground level, at anangle of 66.6° above the horizontal. The rocket is fired towardan 11.0 m high wall, which is located 25.0m away. The rocket attains its launch speed in a negligibly shortperiod of time, after which its engines shut down and the rocketcoasts. By how much does the rocket clear the top of the wall?
1 m

Section 3.2 Equations of Kinematics in twoDimensions
Section 3.3 ProjectileVelocity

Section 3.2 Equations of Kinematics in twoDimensions
Section 3.3 ProjectileVelocity

Explanation / Answer

horizontal velocity=75*cos66.6 horizontal accleration=0 to travel 25 m horizontally the rocket took t s t=25/(75*cos66.6)=0.839 s vertical velocity=75*sin66.6 vertical accleration=g=9.8 time taken to reach max. height =T v=u-g*T 0=75*sin66.6-9.8*T T=7.03 s so the rocket is still climbing when it travelled 25 mhorizontally in 0.839 s the rocket is at a height of h from ground h=u*t-(1/2)*g*t2 h=75*sin53*0.839-(1/2)*9.8*0.839*0.839 h=46.8 m the rocket clear the top of the wall by 36.483-11=35.8 m (ans)

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