Rocket Height Problem: Find y_max w/ 9.8m/s^2

[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.

Please help, I think I'm doing something wrong here, the answer "490 m" doesn't seem correct.

 

[stunner5000pt]

read the question

accelerates straight upward from rest

there is your initial velocity

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.

 

[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)