# Determining Starting Height of Skier

1. Dec 31, 2013

### Softshow

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. Dec 31, 2013

### Gamma

Check the third line from bottom, sqrt(59) - does this make sense?

3. Jan 1, 2014

### Softshow

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?