What acceleration does a rocket need to reach a speed of 240 m/s at a height of 1.2 km?

First you will need this kinematic equation in order to solve this problem:

v(t)^2 = v(i)^2 + 2 *a * x

Final velocity squared = initial velocity squared + 2 x acceleration x distance

First we will convert the distance from 1.2 km into 1200 m by multiplying 1.2 on 1000 . Since that the rocket is moving vertically upward then the final velocity = 0 and the initial velocity = maximum (240 m/s ) and the acceleration will have a negative quantity , so :

0 = (240)^2 + 2 * 1200 * a

- 57600 = 2400 a

a = - 57600 / 2400

then the acceleration = - 24 m/s^2

According to vf(sq = vi(sq + 2ad

vf(sq/2d = a

240(sq / (2 x 1200) =

57600 / 2400 = 24

So it might be 24 m/s/s.

I am only in 9th grade so what would I know :D

initial vel,u=0...final vel.,v=240m/s...height,h=1.2km=1200m.....

nw, v^2=u^2 + 2ah...bt u=0

or accn, a=v^2/2h=24m/s^2