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Kinematic question

  1. Jul 28, 2011 #1
    1. The problem statement, all variables and given/known data


    A water balloon takes .22 s to cross the 130 cm high window, from what height above the top of the window was it dropped?



    3. The attempt at a solution

    I'm using:

    v^2 = vo^2 + 2a(x - xo)

    v = distance/time
    v = 1.30 m / .22 s = 5.91 m/s

    (5.91 m/s)^2 = 0 + 2(9.8 m/s^2)(x)
    x = 1.78 m from release to bottom of window

    1.78 m - 1.3 m = distance of release from top of window

    .48 m


    I'm doing something wrong since the available answers aren't .48m....and it is something simple.
     
  2. jcsd
  3. Jul 28, 2011 #2
    You can't simply use distance divided by time to get the velocity at the bottom of the window, as the balloon is being accelerated due to the earth. What other equation could you apply?
     
  4. Jul 28, 2011 #3
    nevermind i got it.
     
    Last edited: Jul 28, 2011
  5. Jul 28, 2011 #4
    Right, you can solve for v0 from that equation, and use that to obtain the distance above the top of the window the ball was thrown from.
     
  6. Jul 28, 2011 #5
    yeah i see what i did...thanks.
     
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