Question

A 7450 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 500 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?

C. How fast will it be moving just before it crashes?

Answer 1

A)

Consider the journey from the launch pad to the point where the engines fail : initial velocity,

u = 0 m/s

a = 2.25 m/s^2

s = 500 m

v^2 = u^2 + 2as

v^2 = 0 + 2(2.25)(500)

v^2 = 2250 (m/s)^2

Consider the journey from point where the engines fail to the highest point :

initial velocity, u = sqrt(2250) m/s

final velocity, v = 0

acceleration, a = -9.81 m/s^2

v^2 = u^2 + 2as

0 = 2250 + 2(-9.81)s

s = 114.68 m

Maximum height = 500 + 114.68 = 614.68 m

B)

Consider the journey after engine failure before the rocket comes crashing down to the launch pad

initial velocity, u = sqrt(2250) m/s

displacement, s = -500 m/s

acceleration, a = -9.81 m/s^2

s = ut + (1/2)at^2

-500 = (sqrt(2250))t + (1/2)(-9.8)1t^2

4.9t^2 - 47.43t - 500 = 0

Time taken, t = 16 s

C)

Consider the journey after engine failure before the rocket comes crashing down to the launch pad

initial velocity, u = sqrt(2250) m/s

displacement, s = -500 m/s

acceleration, a = -9.81 m/s^2

v^2 = u^ + 2as

v^2 = 2250 + 2(-9.81)(-500)

speed before it crashes, |v| = 109.8 m/s