A 7500 kg rocket blasts off vertically from the launch pad with a constant upwar

Question

A 7500 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s^2 and feels no appreciable air resistance. When it has reached a height of 530 m. its engines suddenly fail so that the only force acting on it is now gravity. You may want to review (pages 50 - 53). For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Up-and-down motion in free fall. What is the maximum height this rocket will reach above the launch pad? y_max = m How much time after engine failure will elapse before the rocket comes crashing down to the launch pad? t = s How fast will it be moving just before it crashes? v = m/s

Explanation / Answer

velocity at the height of 530 m is

v=sqrt(2aS)=49.9 m/s

now if accleration is due to gravity only i.e.9.8 m/s^2

distance travelled=v^2/2g=127.09 m

total height=530+127.09=657.09 m

B

time taken to come from 657.09 m to launch pad =sqrt(2*657.09/9.8)=11.58 sec(using S=ut+0.5at^2)

time taken to go from launch pad to 530 m =21.24 sec(using S=ut+0.5at^2)

time taken to go from 530 m to 657.09 m =5.09 sec(using S=ut+0.5at^2 here S=127.09,u=0,a=9.8 so we will find t)

total time=37.9 sec

but engine failed at 530 m so time taken after engine failure=11.58+5.09=16.67 sec

C

velocity=9.8*11.58=113.5 m/s

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