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Determining Starting Height of Skier

  1. Dec 31, 2013 #1
    1. The problem statement, all variables and given/known data

    A skier starts from rest at height H on the side of a valley. The other side of the valley rises to a rounded peak at 4.0m. Assuming the skier goes down one side of the valley and up the other, what is the maximum value of H for which the skier remains in contact with the ground at the peak of the far side?

    2. Relevant equations
    Minimum velocity at top of far side = √(yg)
    1/2(mv^2)+mgy=mgH

    3. The attempt at a solution
    Vmintop = √(4.0m(9.8m/s^2))
    Vmintop = √(39.2m^2/s^2)
    Vmintop = 6.3m/s

    1/2mv^2+mgy = mgH
    1/2v^2+gy = gH
    1/2(6.3m/s^2)^2+(9.8m/s^2)(4.0m) = 9.8m/s^2(H)
    20m^2/s^2+39m^2/s^2=9.8m/s^2(H)
    20m^2+39m^2 = 9.8m(H)
    √(59m^2) = 9.8m(H)
    7.7m = 9.8m(H)
    H=0.79m

    Not sure where I went wrong, may be as simple as an arithmetic error, but I can't find it for the life of me.
     
    Last edited: Dec 31, 2013
  2. jcsd
  3. Dec 31, 2013 #2
    Check the third line from bottom, sqrt(59) - does this make sense?
     
  4. Jan 1, 2014 #3
    Oh I see, because if I was to divide 7.7m by 9.8m i wouldn't be left with a unit. Instead, I suppose I should just divide 59m^2 by 9.8m, making H=6.0m?
     
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