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Kinematics Problem on toboggan

  1. Nov 1, 2007 #1
    1. The problem statement, all variables and given/known data
    At the base of a frictionless icy hill that rises at 28.0 degrees above the horizontal, a toboggan has a speed of 12.0m/s toward the hill. How high vertically above the base will it go before stopping?


    2. Relevant equations
    height = [(v*sin (angle))^2] / (2g)


    3. The attempt at a solution
    height = [(v*sin (angle))^2] / (2g)
    height = [(12m/s*sin(28degrees))^2] / (2(9.8m/s^2))
    height = 31.74m^2/s^2 / 19.6m/s^2
    height = approx 1.62m

    I think I made a mistake with the velocity because this answer is incorrect. Can someone help me please and thank you.
     
  2. jcsd
  3. Nov 1, 2007 #2

    Dick

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    I don't see any problem with that. I get the same thing.
     
  4. Nov 1, 2007 #3
    I would do this problem with a more general equation like the conservation of energy (KE + PE = Etotal). It looks like the equation you gave was something that you didn't derive yourself and was specific to another situation. If you did derive this equation, rethink the velocity.
     
  5. Nov 1, 2007 #4
    No, I disagree with his current answer. Dick, when using the conservation of energy, it doesn't matter which direction v is in. So there's no need to find a specific component of v.
     
  6. Nov 1, 2007 #5
    In that case...

    height = (v^2) / (2g)
    height = (12m/s)^2 / (2*9.8m/s^2)
    height = 144m^2/s^2 / 19.6m/s^2
    height = 7.35m

    or if I used energy...

    KE=GPE
    0.5mv^2 = mgh
    0.5v^2 = gh (mass drops out)
    0.5*(12m/s)^2 = h*9.8m/s^2
    72m^2/s^2 = h*9.8m/s^2
    (72m^2/s^2) / (9.8m/s^2) = 7.35m

    This answer worked. Thanks Anadyne. Can't get em all Dick ;p
     
    Last edited: Nov 1, 2007
  7. Nov 1, 2007 #6

    Dick

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    Ooops. Sorry. Guess it's time to lay off the problem solving for tonite. Thanks for the correction.
     
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