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anonymous

  • 5 years ago

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8 m/2^2. The acceleration period lasts for time 8.00 S, until the fuel is exhausted. After that, the rocket is in free fall. Find the maximum height Vmax reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.80 m/s^2 .

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  1. anonymous
    • 5 years ago
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    *58.8 m/s^2

  2. anonymous
    • 5 years ago
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    i found the height when the fuel is exhausted which is = to 1880

  3. anonymous
    • 5 years ago
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    the velocity at that point i think is 470 m/s

  4. anonymous
    • 5 years ago
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    i think i have to find the distance it went after that and add it?

  5. anonymous
    • 5 years ago
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    yep

  6. anonymous
    • 5 years ago
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    that 1880, how did you come to that

  7. anonymous
    • 5 years ago
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    1881.6 lol

  8. anonymous
    • 5 years ago
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    yea rounded wrong LOL

  9. anonymous
    • 5 years ago
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    470.4 m/s correct

  10. anonymous
    • 5 years ago
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    I must complain about this question - it doesn't take into account the loss of mass as the fuel is ejected.

  11. anonymous
    • 5 years ago
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    yea physics is a pain.. LOL

  12. anonymous
    • 5 years ago
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    hey man it's simple physics lol. if you take mass into account, school students are doomed

  13. anonymous
    • 5 years ago
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    i was using this formula Xf=Xi+(Vx)i Dt + 1/2Ax(Dt)^2

  14. anonymous
    • 5 years ago
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    when ever i see a physics problems i get anxiety.. =(

  15. anonymous
    • 5 years ago
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    ttt luv the pic

  16. anonymous
    • 5 years ago
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    anyway, what's the distance traveled after fuel ran out?

  17. anonymous
    • 5 years ago
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    i couldnt figure it out

  18. anonymous
    • 5 years ago
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    im thinking i have to use this formula V^2 = Vo^2+2a(y-yo) Vo is the initial velocity which is 470 m/s?

  19. anonymous
    • 5 years ago
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    ok, take upwards as possitive direction at the point where feul ran out, u=470.6m/s, a=-9.8m/s^2m, v=0m/s at max point. use v^2=u^2+2aS

  20. anonymous
    • 5 years ago
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    a=-9.8m/s^2 negative since acceleration downwards. gravity

  21. anonymous
    • 5 years ago
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    uh huh!

  22. anonymous
    • 5 years ago
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    so plug them in the equation, you'll find distance from no more gas point to max point, add this to distance from ground to ran out point to get max height

  23. anonymous
    • 5 years ago
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    what do i plug in for S?

  24. anonymous
    • 5 years ago
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    that's what you need to find. plug everything in to find S

  25. anonymous
    • 5 years ago
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    0=470.6^2 + 2 (-9.8)S

  26. anonymous
    • 5 years ago
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    yeap

  27. anonymous
    • 5 years ago
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    11299

  28. anonymous
    • 5 years ago
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    yea, looks reasonable add this S to the S before(1881.6) you get max height

  29. anonymous
    • 5 years ago
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    damn man youre a genius!

  30. anonymous
    • 5 years ago
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    it took me an hour to figure out the first part =(

  31. anonymous
    • 5 years ago
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    don t worry you will get used to it very soon, just keep practicing. All those problems more or less are the same

  32. anonymous
    • 5 years ago
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    thanks! when can i buy you lunch!? LOL

  33. anonymous
    • 5 years ago
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    lol what do you think, I eat physics for lunch hahaha. just kidding. I'm off to bed now. see ya.

  34. anonymous
    • 5 years ago
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    haha thanks good night!

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