Set Upward Accel. Then FreeFall ?

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The discussion revolves around a physics problem involving a rocket that accelerates upward before free-falling back to Earth. The key points include the need to determine the maximum height the rocket reaches after its engines shut off, with the understanding that its velocity will be zero at that peak. The upward motion is divided into two phases: the powered ascent and the subsequent ascent influenced only by gravity. To solve for the maximum height, the correct equations of motion must be applied, particularly focusing on the transition from powered flight to free fall. The conversation emphasizes the importance of correctly applying the kinematic equations to find the distance traveled during the rocket's ascent after engine cut-off.
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Homework Statement


For some reason, I just can't seem to understand this question. I didn't think it would be this bad. Anyways, A Rocket fires from rest with an upward acceleartion of 30 m/s^s for 2 seconds. After this time the engine shuts off and the rocker freely falls back to the surface of the Earth. ( use g = 10 m/2^s)
a) Draw a v vs t graph. I've done that already. ;'D
b) Determine the height of the rocket when the engines shut off. Done.
c) Determine the Maximum height that the rocket reaches.<--- Okay. THis is the one. <--- For some reason I'm thinking that at the crest of the curve/line/ place before it falls its V = 0, but I have no idea how to figure out how much farther/higher it goes. And that is my problem.



2. The attempt at a solution

I've tried using, x = 1/2at^2+VoT+Xo and V^2 = Vo2+2AX.
But neither are right. Or I'm doing something wrong. D:
THanks in advance for any help or comments. :'D

Homework Statement

 
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You're right in saying that the at the crest of its trajectory the rocket's velocity is zero.

The rocket's upward motion can be split in two parts, 1. The part with the thrust and 2. The part where its still moving up with only the influence of gravity. For the second part, the acceleration is known (what are the forces acting on it?), and so is the initial velocity (look at the graph) and final velocity. Now use "V^2 = Vo2+2AX" to find the distance traveled during the second part of it's upward journey.
 
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