Question

a rocket takes off vertically from rest under the power of its rocket motor. The rocket motor is sufficiently powerful that the rocket has has an upward acceleration of 5g. after 5 sec the rocket motor instantaneously burns out and only the downward acceleration of gravity acts on the rocket. Approximately what maximum height does the rocket reach above the point where its motor burns out

Please show work so I know how to solve the problem myself for future problems

Answer 1

The speed of the rocket when it’s motor burns out:
v=at=5g(5)=25g.
From the conservation of energy:
(mv^2)/2=mgh
The maximum height does the rocket reach above the point where its motor burns out:
h=v^2/2g=(25g)^2/2g=625/2 g=625/2 (9.8)≈3000 m=3 km.