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Speed up a hill

  1. Feb 7, 2008 #1
    1. The problem statement, all variables and given/known data
    At the base of a frictionless icy hill that rises at 29.0deg above the horizontal, a toboggan has a speed of 12.0 m/s toward the hill. How high vertically above the base will it go before stopping?


    2. Relevant equations
    Wtot=K2-K1
    K=1/2mv^2
    K2=1/2mv^2=K1+Wtot


    3. The attempt at a solution
    Vx=12cos(29) = 10.5 m/s
    vf^2-vi^2=2as
    0-(10.5)^2=2(-9.8)s
    s= 5.625
     
  2. jcsd
  3. Feb 7, 2008 #2
    This is how i would do it:
    1. draw an fbd of the toboggan
    2. use f=ma to solve for a
    3. solve for distance travelled on the hill
    4. then you make a triangle, with the distance you found in 3) being the hypotenuse, solve for the vertical side, and thats how high it has travelled.
     
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