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.