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Constant vertical acceleration

  1. Jan 25, 2008 #1
    1. The problem statement, all variables and given/known data

    A 7400 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.15 m/s^2 and feels no appreciable air resistance. When it has reached a height of 560 m, its engines suddenly fail so that the only force acting on it is now gravity. a. What is the maximum height this rocket will reach above the launch pad? b. How much time after engine failure will elapse before the rocket comes crashing down to the launch pad? c. How fast will it be moving just before it crashes?

    2. Relevant equations

    560=.5(2.15)t^2.
    v0=2.15t
    ht=560Tv0t-.5(2.15)t^2

    3. The attempt at a solution

    I got 1115 meters and it's wrong.
     
  2. jcsd
  3. Jan 25, 2008 #2

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    OK, but not necessary. You have to find the velo vf at that height. Use the other formula.

    ??

    Use vf as initial velo at that height and find max height. The other answers will follow.
     
  4. Jan 25, 2008 #3
    How do I find t?
     
  5. Jan 25, 2008 #4
    how do I find final velocity when just given acceleration and height?
     
  6. Jan 25, 2008 #5

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    Let me use some symbols. The rocket is at a height h when the engines stop. The velo is vf at that point. It rises to H before it falls back. I'm denoting the height of the rocket at any time by y.

    1. You can find H-h if you know vf and g. Use formula for final, initial velos and height.

    2. From h, you can find t for y to become zero, that is, time to fall. Use directly y = ut-1/2 gt^2. Decide on what you'll put as y.
     
  7. Jan 25, 2008 #6

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    Vf^2 = Vi^2 + 2ah. This is when the rocket is rising with const accn 'a', OK?
     
    Last edited: Jan 25, 2008
  8. Jan 25, 2008 #7
    What does u represent?
     
  9. Jan 25, 2008 #8

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    The initial velo, that is, vi. Have you understood the rest?
     
  10. Jan 25, 2008 #9
    No. I have been looking at this for five hours now. I don't know the velocity or the time. So I don't know how I can solve the problem.
    ,
     
  11. Jan 25, 2008 #10
    does final velocity equal 2 ah?
     
  12. Jan 25, 2008 #11

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    What does post #6 say?
     
  13. Jan 25, 2008 #12
    final velocity^2 +initial velocity^2 +2ah.

    I dont know the final velocity or the initial velocity.
     
  14. Jan 25, 2008 #13
    All these formulas have two unknowns.
     
  15. Jan 25, 2008 #14
    I know when the rocket hits its maximum height, the velocity is zero.
     
  16. Jan 25, 2008 #15

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    final velocity^2 =initial velocity^2 +2ah.

    vi=0, a=2.15, h=560. What more do you want to find vf?

    (Nothing has been said about the initial velo of the rocket, so it is assumed to be zero, if that's what was stopping you. Anyway, a rocket is not thrown upward, it rises from rest due to accn.)
     
  17. Jan 25, 2008 #16
    The final velocity is 49 m/s Now what do I do?
     
  18. Jan 25, 2008 #17
    I got 881 for height.
     
  19. Jan 25, 2008 #18
    I got 650 and it says there is a rounding error.
     
  20. Jan 25, 2008 #19

    hage567

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    So 650 is another attempt at finding the height?? You're going to have to show more detail of your calculations for us to spot where you might have gone wrong.
     
  21. Jan 26, 2008 #20

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    Correct.

    Decide on one, and show your calc. Incidentally, neither is correct.

    Try to understand. At the point where the engines switch off, the rocket is moving under g only. It's initial velo is 49 m/s upward. So, to what height does it rise from that point? Final velo will be zero, and that's how you find the height. So, you know total height H.

    For finding the time to fall to ground, use the height h=560 m as your starting posn. Use y=vi*t-1/2 gt^2. Then what will be the 'y' for the ground? What will be vi?

    (Gather your thoughts and send fewer posts, instead of writing a line and posting it.)
     
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