SITUATION: A 7500-kg rocket blasts off vertically from the launch pad with a con
ID: 2025242 • Letter: S
Question
SITUATION: A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25 m/s2 and feels no appreciable air resistance. When it has reached a height of 525 m, its engines suddenly fail so that the only force acting on it is now gravity.a) What is the maximum height this rocket will reach above the launch pad?
b) How much time after engine failure will elapse before the rocket comes crashing down to the launch pad?
Use a detailed diagram to explain the above.
A diagram is very important.
Thank you!
Explanation / Answer
2.a) Use the kinematic equation, to find the final velocity when the rocket engine stops. v^2 = u^2 + 2 a s u = 0 a = 2.25 ms^2 s = 525 m v = 48.6 m/s Now with initial velocity v the maximum height reached h is again found with the equation , 0 = v^2 - 2 g h h = v^2 / 2 g = 48.6^2 / 2 x 9.8 = 120.5 m The total height reached by the rocket is H = h + s = 645.5 m b) Time taken to reach the maximum height after the engine shuts down, is found by 0 = v - g t' t' = v / g = 48.6 / 9.8 = 4.96 s Time taken for the free fall from a height of 645.5 m is H = 0 + g t''^2 /2 t'' = sqrt( 2 x H / g ) = 11.48 s Time elapsed after the engine failure before the rocket crashes is t = t' + t'' = 16.44 s Final velocity with which the rocket crashes is given by v' = 0 + g t'' = 112.48 m/s
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.