Physics question based on acceleration?

Question

A rocket accellerates at 5.00m/s until it runs out of water 2.5s later.How high will the rocket be when it's velocity reaches zero?

Answer 1

The problem should read 5.00 m/s^2 but assuming that is only a typing error,
we have to use one of the laws of motion which is best suited for the purpose :
Here we know
u = initial velocity = 0 m/s
a = acceleration = 5 m/s^2
t = time = 2.5s
first we find the velocity after 2.5 s or final velocity v
v = u + at
= 0 + 5 * 2.5 = 12.5 m/s
now we have to consider that after the fuel (water?) is finished the rocket will experience gravity at 9.8 m/s^2 in the downward direction and will decelerate :
now the initial velocity of the rocket is the final velocity we had calculated earlier or 12.5m/s
its final velocity is 0 m/s
deceleration = 9.8 m/s^2
time taken = (v - u) / a
= ( 0 - 12.5) / -9.8 (the negative sign is because the acceleration is in the opposite direction.)
= 1.28 s
now we calculate the distance it travels in these two stages.
stage 1
s = ut + 0.5 a t^2
= 0 + 0.5 * 5 * 2.5^2
= 15.625 m
stage 2
s = ut - 0.5 a t^2
= 12.5 *1.28 - 0.5 * 9.8 *1.28 ^2
= 16 - 8.03 = 7.97 m
total distance travelled =
15.625 + 7.97 = 23.595m --------- answer.

Answer 2

v = 2.5(9.8-5) = 12 m/s

S = 0.5(9.8-5)*2.5*2.5

S1 = 12*12/2*9.8

Add S and S1