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Rocket height problem

  1. Sep 27, 2007 #1
    1. The problem statement, all variables and given/known data
    A rocket moves upward, starting from rest with an acceleration of 31.0 m/s2 for 5.00 s. It runs out of fuel at the end of the 5.00 s but does not stop. How high does it rise above the ground?

    2. Relevant equations

    x=1/2at^2

    3. The attempt at a solution

    x=1/2(31.0)(5.00)^2
    I got x=388 m
     
  2. jcsd
  3. Sep 27, 2007 #2
    I'm assuming that's not the solution. The distance you got was the distance traveled from rest to the point at which there was no more acceleration.

    The next step is to find the distance after the fuel is out.
    Vo=velocity at the point when fuel runs out. You would use vf^2=vo^2+2ad. vo is given. vf is implied. a is also implied. Find d.
     
  4. Sep 27, 2007 #3

    D H

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    That's how far it has risen at the point it runs out of fuel. You were told explicitly that the rocket doesn't stop just because it runs out of fuel. What happens after it runs out of fuel?
     
  5. Sep 27, 2007 #4
    DH;
    i guess then it would start descending due to the pull of gravity? i don't understand what silvashadow is trying to tell me. i'm a bit confused.
     
  6. Sep 27, 2007 #5
    Yes, gravity is the only thing decelerating it. Air resistance is neglected because this is beginning physics, the same part I'm learning now. There are some equations that relate velocity to time and distance. vf^2=vo^2+a2ad is an equation.

    You could use the basic vf=vo+at to fine the time it takes. Then you can use d=.5(vf+vo)t to find the distance traveled.
     
  7. Sep 27, 2007 #6

    D H

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    When you toss a rock up in the air does it stop going up as soon as it leaves your hand or does it fly up in the air a bit after leaving your habd before it stops going up? Why would the rocket behave any differently?
     
  8. Sep 27, 2007 #7
    yes i did use that equation and i still have 388 m. i must be putting in a wrong number for something. lets see i did Vf=0+31.0*5.00 and got 155m/s. so then i used x=1/2(Vf+Vo)t or x=1/2(155+0)5.00 and got 387.5 or 388 m.
     
  9. Sep 27, 2007 #8

    D H

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    You are still finding the height at which the rocket engine stops firing, which is not the height the rocket stops going up. You now have that the velocity when the rocket stops firing is 155m/s upward. What happens from this point on?
     
  10. Sep 27, 2007 #9
    distance traveled by rockets:x=.5at^2, a is 31, given, t is 5, given
    distance traveled w/o rockets: vf^2=vo^2+2ad, a=-9.8, vo=at=(31)(5), vf=0, solve for d
     
  11. Sep 27, 2007 #10
    oh ok. i found the distance for both and added them. I got 1612 m but i changed the sig figs for the problem and ended up with 1.61e3? right?
     
  12. Sep 27, 2007 #11
    you must have rounded, i got 1613, but yes 1613 = 1.613e3
     
  13. Sep 27, 2007 #12
    THANK YOU SOOO MUCH!!!YOU BASICLY SAVED MY LIFE!yes that was the correct answer. i finally get it though it's the distance it went, then once it stopped the distance it went a little more because it didn't actually stop all the way and the total which was how far it was off the ground all together. i don't think i said that clearly but yeah i get it now. THANK YOU!!!
     
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