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Homework Statement
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?
Homework Equations
Minimum velocity at top of far side = √(yg)
1/2(mv^2)+mgy=mgH
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.
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