NEED HELP>..celestial mechanics...have the correct answer?

Question

a 2300kg rocket shuts off its engine 494km above earths surface.its velocity at burnout is 3000m/s directly upward.ignoring air resistence, what maximum height will the rocket reach?
the answer is 1.1 x 10^6 metres
thx!!

Answer 1

v^2 - u^2 = 2as is the formula to be used.

v = 0 when the rocket stops climbing

u = 3000 m/sec

a = -g (acceleration due to gravity)

s = distance travelled

s = u^2 / 2g = 3000 x 3000 / (2 x 10) (let us assume g is 10m/s^2)

s = 4.5 x 100000 = 450000 m

Already it has reached a distance of 450km and so the total will be 900,000m.

We made an assumption that g will be constant which is not so. We need to caculate the variation of g with distance and use the average g from the point 450 km above earth to the final resting point and since g does reduce with distance, the distance reached will be more.

I am not doing those calculations for now but that can be done.