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Rocket with constant acceleration - am i right or not ?

  1. Sep 8, 2011 #1
    1. The problem statement, all variables and given/known data

    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?


    2. Relevant equations

    [tex]s(displacement)=v_0t + \frac 12 at^2[/tex]

    3. 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!
     
  2. jcsd
  3. Sep 8, 2011 #2

    Janus

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    Gold Member

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

    Secondly, what happens to the rocket after its engines stop firing?
     
  4. Sep 8, 2011 #3
    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?
     
  5. Sep 8, 2011 #4
    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?
     
  6. Sep 8, 2011 #5
    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...
     
  7. Sep 8, 2011 #6
    Yes, that was the way.

    Many thanks for your help, guys !!!
     
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