# Rocket Height problem.

1. Feb 2, 2006

### DefaultName

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8 m/s^2. The acceleration period lasts for time 10.0 s until the fuel is exhausted. After that, the rocket is in free fall.

Find the maximum height y_max reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.8 m/s^2.

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The way I tackled this problem was I used the equation:

final position = intial position + vt + 1/2(a)(t^2)

I got an answer of 490m, but that doesn't seem to be correct. I'm not sure if I have to use the acceleration given to me in the equation (58.8m/s^2) or the default acceleration due to gravity = 9.8 m/s^2.

This answer (490 m) shows how far the rocket would fall in time t_1 starting from zero velocity. I'm not sure if there's an intial velocity (the problem doesn't state it), but if there is, I'd probably use vf^2 = vi^2 + 2as.

Last edited: Feb 2, 2006
2. Feb 2, 2006

### stunner5000pt

what is the NET acceleration on the rocket? It accelerates upward with 58.8 and gravity pulls it downward at 9.8. What is the NET acceleration as a result?

you are thus given intial velocity, time, net acceleration, can you find the distance (max height) reached.

3. Feb 2, 2006

### DefaultName

Hmm..

I tried doing it with my acceleration as 49m/s^2, but my online system is saying no... it specifically says

"Read the problem carefully; a = 58.8 m/s^2 is the net acceleration of the rocket."

How does that work?

I'm using the equation final height = intial height + vt + 1/2at^2
where I need to find final height = 0 + 0 + 1/2 (??) (100)

Last edited: Feb 2, 2006