Work energy theorem questions

please give the solution to the following:
Q.1
if you have a rocket with a mass of 60 t, force of 2.50 MN and will fire its engine for the time of 10 s. for how long will the rocket continue to climb and what Maximum height will it reach before falling back if you have the height of 2 km and speed of 0.40 km/s at the moment the engine stops? ans: t=40, d=2 km

Q.2
to lift off the surface of the moon the astronauts used a lunar module of average mass of 2.5 t and 4 GJ of energy. to rejoin the command module, the astronauts must have a speed of 1.6 km/s. The gravity on the moon is 1.6 m/s^2. Find the height of command module. Ans: 200km

Thanks in advance! I've tried many times but i can't get it right...

Q1 . The answer you give for max height makes no sense. If the rocket is at a height of 2km and is travelling upwards at 400m/s when the engine stops, obviously the maximum height must be more than 2km.

Your answers seem to assume g=-10m/s², so that's what I'll use.

From the moment when the engine stops (when u=400m/s) to the max height (when v=0) we can use:
v² = u² + 2as (or vf² = vi² + 2ax if you use these symbols)
0² = 400² + 2(-10)s
s = 8000m = 8km

Since the engine stopped when the height was 2km, the max height = 2 + 8 = 10km

When the engine stopped the rocket continued to rise for a time t where:
v = u + at
0 = 400 + (-10)t
t =40s
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Q2 The required kinetic energy of the module =½mv² =½ x 2500 x (1600)² = 3.2x10^9J
Total energy needed = potential energy gained + kinetic energy needed
4x10^9 = PE + 3.2x10^9
PE = 8x10^8 J

Assuming 'g' is constant over the height change, and using PE = mgh
8x10^8 = 2500 x 1.6 x h
h = 2x10^5m = 200km