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Homework Help: Vertical Drop how to find the speed?

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data

    The Tower of Doom at Six Flags America lifts its riders to a height of 42.7 m, then releases them to free fall for 21.6m. How fast are the riders traveling before the magnetic induction braking system starts to slow them down?


    2. Relevant equations

    t = [tex]\sqrt{}2d/g[/tex]

    3. The attempt at a solution

    t = (2)(21.6) / 9.81, square root
    t = 4.40, square root
    t = 2.10 s

    21.6m / 2.10s
    Speed = 10.3 m/s

    I just wanted to see if I actually did this right and if I didn't can someone help me with it?

    P.S. when i say ", square root" that means there's a square root in the equation
     
  2. jcsd
  3. Jul 11, 2010 #2

    Pengwuino

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    No that is not right. You calculated the average speed. If the riders drop with an acceleration of 9.8m/s^2 for a certain time t, what does that tell you about their acceleration at that time t?
     
  4. Jul 11, 2010 #3
    their acceleration at 2.10s x 9.81m/s/s would equal to 20.6m/s/s/s
    is it wrong because there are too many seconds?
     
  5. Jul 11, 2010 #4

    Pengwuino

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    No, the seconds in the time will cancel one of the seconds in the acceleration. Also, upon further inspection (ok, first inspection :redface: ), you are using the wrong distance. You need to use the actual displacement, 42.7m - 21.6m = 21.1m. Also, just to be a little nit-picky, to 3 significant figures, g = 9.80m/s^2.
     
  6. Jul 11, 2010 #5
    oh, wow I thought the displacement would be 21.6 because that's how long the drop down was. Huh, okay so wait should I try the same formula again but with the 21.1m?

    If I did use the same formula would it be:
    (2)(21.1) / 9.80 square root
    t = 2.08 s?
     
  7. Jul 11, 2010 #6

    Pengwuino

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    Bah, sorry! I misread the problem! Yes, you're right, 21.6. I thought it said they drop TO 21.6m
     
  8. Jul 11, 2010 #7
    omg i actually did do something by myself right >.<
    but wait wait is the formula still right to find time?
     
  9. Jul 11, 2010 #8

    Pengwuino

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    Yes, given an acceleration with no initial velocity and a displacement, d, that is how you find the time required to achieve that displacement.
     
  10. Jul 11, 2010 #9
    so then time would equal to 2.1 s
    and then
    21.6m / 2.10s
    Speed = 10.3 m/s

    :biggrin: please tell me I'm right :uhh:
     
  11. Jul 11, 2010 #10

    Pengwuino

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    No no, you had it right at post #3 except for the minor g = 9.81 or g = 9.80 thing. The velocity is just Velocity = Acceleration * Time for a constant acceleration.
     
  12. Jul 11, 2010 #11
    ohhh okay thank you so much!!!! :!!):biggrin::!!):biggrin:
     
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