Physics help.?

Question

I need help with this physics problem. I have been trying to solve this problem for 2 hours but I can't find the right answer.
Here's the problem:

A rocket moves upward, starting from rest with acceleration of + 29.4 m/s^2 for 3.98s. It runs out of fuel at the end of the 3.98 s but does not stop. How high does it rise above the ground?

In the back of the book the answer is 931 meters, but I want ot know how to get there.

Thank you for your help!!!!

Answer 1

well there are two parts to the rockets travel.
the first part with fuel is average velocity times time
d = (at)/2 x t, commonly written 1/2 at^2
d = .5 x 29.4 x 3.98 x 3.98
that gives how high it goes on fuel

after that it is a free falling object. Its initial upward velocity = at = 29.4 *3.98
it goes up for some time related to g. when g x t = the initial velocity you know how long it continued going up.
given that time you can calc the distance as average velocity times time, 1/2 v x t

so then add the two stages. I get 931...