Set Upward Accel. Then FreeFall ?

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SUMMARY

The discussion focuses on a physics problem involving a rocket that accelerates upward at 30 m/s² for 2 seconds before free-falling back to Earth. The key points include determining the maximum height reached by the rocket after the engine shuts off, where the velocity at the peak is zero. The correct approach involves using the kinematic equation V² = Vo² + 2AX to calculate the additional height gained during the free ascent after engine cutoff.

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  • Understanding of kinematic equations in physics
  • Knowledge of acceleration due to gravity (g = 10 m/s²)
  • Ability to interpret velocity vs. time graphs
  • Familiarity with concepts of upward and downward motion
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  • Explore the effects of gravity on free-falling objects
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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+volt+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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