A rocket starts from rest at the earth\'s surface and accelerates straight upwar

Question

A rocket starts from rest at the earth's surface and accelerates straight upward with a constant upward acceleration of 20 m/s^2. At a height of 800 m the engine cuts off and the rocket continues to move upward in free fall, reaches a maximum height, and then falls back down to earth. (Air resistance can be neglected)

a. What is the speed of the rocket at the instant when the engines cut off?

b. what is the maximum height above the ground reached by the rocket?

c. What is the speed of the rocket just before it strikes the ground?

Explanation / Answer

a) v^2 - u^2 =2ah

v^2 - 0 = 2 x 20 x 800

v =178.89 m/s

b) after 800 m

h = v^2/2g = 178.89^2/2x9.8 =1632.65 m

H = 800 + 1632.65 = 2432.65 m

c) v^2 - u^2 =2ah

v^2 - 178.89^2 = 2 x 9.8 x 800

v =218.36 m/s downwards

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