A 7800 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.20 m/s2 and feels no appreciable air resistance. When it has reached a height of 595 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?

During the blast off, the rocket's upward acceleration is constant at 2.20 m/s². When the engines fail and the only force acting on the rocket is gravity, it's acceleration becomes -9.8 m/s². We can use constant acceleration kinematics (also known as SUVAT) to find the maximum height, time, and speed.

a) During the first 595 m, the rocket accelerates upward at 2.20 m/s². The speed it reaches at this point is:

v² = u² + 2as

v² = (0 m/s)² + 2 (2.20 m/s²) (595 m)

v = 51.2 m/s

The engines fail, and the rocket now accelerates at -9.8 m/s². At the maximum height, the speed is 0 m/s. The distance the rocket travels during this time is:

v² = u² + 2as

(0 m/s)² = (51.2 m/s)² + 2 (-9.8 m/s²) s

s = 134 m

The total height is therefore:

595 m + 134 m = 729 m

b) When the engines fail, the rocket will rise to its maximum height, then fall back to the ground, 595 meters below the point where the engines failed. The time it spends in free fall is:

s = ut + ½ at²

-595 m = (51.2 m/s) t + ½ (-9.8 m/s²) t²

4.9t² − 51.2t − 595 = 0

t = [ -(-51.2) ± √((-51.2)² − 4(4.9)(-595)) ] / 2(4.9)

t = (51.2 ± 119.5) / 9.8

t = 17.4 s

c) The speed at which it hits the ground after falling from its maximum height is:

v² = u² + 2as

v² = (0 m/s)² + 2 (-9.8 m/s²) (-729 m)

v = 119.5 m/s