Rocket with constant acceleration - am i right or not ?

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Homework Help Overview

The problem involves a rocket launched vertically with a constant acceleration of 20 m/s² for 1 minute, followed by free fall after the fuel is exhausted. The original poster seeks to determine the maximum height reached by the rocket.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the height using the displacement formula, questioning the validity of their initial speed assumption. Other participants point out a numerical error and inquire about the rocket's behavior after engine cutoff, leading to discussions about the second phase of motion and the need for additional calculations.

Discussion Status

The discussion is active, with participants exploring the dynamics of the rocket's ascent and subsequent free fall. Guidance has been provided regarding the need to calculate the final velocity at the end of the powered ascent to determine the initial conditions for the free fall phase.

Contextual Notes

There is a discrepancy between the original poster's calculation and the answer key, prompting a review of assumptions and calculations. Participants are also addressing the transition from powered flight to free fall, which introduces additional considerations for the problem.

mstud
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Homework Statement



A rocket is fired vertically and contiues for 1 minute with the acceleration 20 m/s^2. Then it has used up all the fuel , and the rocket continues in free fall.

a) How high does the rocket come?


Homework Equations



s(displacement)=v_0t + \frac 12 at^2

The Attempt at a Solution



I can also assume that the initial speed v0 is = 0 m/s, right?

This gives s= 0*60s + 1/2 * 20 m/s^2 * (60s)^2 = 36000 m =3.6*10^5 m.

However, the answer key of my book says s=1.5*10^5m.

Can you find something wrong in what I have done or is my book wrong?

Please help!
 
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First off. 36,000 is 3.6*10^4.

Secondly, what happens to the rocket after its engines stop firing?
 
Yes, of course, sorry ...

Continues with the acceleration -9.81 m/s due to gravity, until v=0 m/s, right?
How to use that?
 
You now have a second "leg" (as it were) of the journey. You have

a final velocity (at peak height)
an acceleration due to gravity alone

Take a look at your equations of motion and see if any of them use these. You'll find that you need one more item from one of the equations, the initial velocity. How can you find the initial velocity of the second leg of the journey?
 
The initial velocity of the second leg is equal to the final velocity of the first leg, so I can calculate the final velocity of the first leg to find it :)

Then the total displacement is first leg + second leg.

I'll see if this solves it...
 
Yes, that was the way.

Many thanks for your help, guys !
 

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