Physics Help - Please :D


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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 29.4ms^-2. The acceleration period lasts for time 6.00s until the fuel is exhausted. After that, the rocket is in free fall. Find the maximum height reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.8ms^-2.

A very simple question, but I can't seem to get the answer. Any help would be greatly appreciated :)

I know you've to add 2 values for the maximum height. Can't seem to get the correct numerical answer though.

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I haven't done Physics for a long time, so I may have made a stupid mistake but here's my attempt :p

// part 1

u = 0, a = 29.4ms^-2, t = 6

s = ut + 1/2at^2 = (0.5 * 29.4 * 6^2) = 529.2m

v = u + at = 29.4 * 6 = 176.4ms^-1

//part 2

u = v1 = 176.4ms^-1, v = 0, a = -9.8

v^2 = u^2 + 2as => s = (v^2 - u^2)/2a = (-176.4^2)/(-9.8 * 2) = 1587.6m

final answer; s1 + s2 = 529.2 + 1587.6 = 2116.8m

Edited by ZakO
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what zako has done is correct, here's the explaination. Basically you split this problem into two parts. First part using your initial velocity as 0 and final velocity as 'x' (unknown). using the equation s=ut+1/2at^2 * v=u +at you can calculate the final velocity & the distance covered in P1 (part 1).

for the second part you take the final velocity of p1 and use it as the initial velocity for p2 and since you know the rocket is going to be at a standstill at the maximum point you take the final velocity to be 0. in this part use acceleration as the value of gravity (but -ve because gravity is pulling the body back to earth). again, find the distance covered. add distance from p1+p2 and you get your answer.

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