Physics problem involving a skier on a hill. I've spent a lot of time on this, but can't figure it out. Help!?

Question

A skier starts from rest at the top of a large hemispherical hill of radius R. Neglecting friction, show that the skier will become airborne at a distance of h=R/3 below the top of the hill.

Anonymous · Accepted Answer

Difficult to show without a diagram but here goes. The skier will leave the hill when the Normal force along the radius=0. At that point the only force along the radius is the vector component of gravity along the radius.  Let's say that point is an angle A from the top of the hill as seen from the radius of the hemisphere.Since just before the skier reaches that point N2nd for the radial motion this situation is 

netforce = mv^2/R (centripetal acceleration)

netforce = mgcosA (vector component of gravity along the radius)
so mgcosA=mv^2/R

note: cosA = (R-h)/R when the skier drops a distance h.

-------> mg(R-h)/R = mv^2/R
-------> g(R-h) = v^2
by conservation of energy loss of gpe=gain in KE

mgh= 1/2mv^2 -------> v^2 =2gh

-------> g(R-h) = 2gh
-------> R-h = 2h
-------> R = 3h

--------> h = R/3 QED

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Physics problem involving a skier on a hill. I've spent a lot of time on this, but can't figure it out. Help!?

1 Answer