A skier starts at rest at the top of a large hemispherical hill (Fig. P7.63). Neglecting friction, show that the skier will leave the hill and become airborne at a distance h 5 R/3 below the top of the hill. Hint: At this point, the normal force goes to zero.

Explanation:

Let at a height h  below the top , normal reaction of surface becomes zero and skier becomes airborne . Let his position is making angle θ with the vertical . Let the radius of the hemisphere be R .

Then

h = R - R cosθ

At the point where the skier becomes airborne

Centripetal force = component of his weight towards the centre of the hemisphere

mv² / R = mg cosθ , m is mass of the skier , v is his velocity at the time of being airborne.

v² = 2gh

= 2gR(1 - cosθ )

Putting this equation in the equation above

m 2gR(1 - cosθ ) / R = mg cosθ

2- 2 cosθ =  cosθ

cosθ = 2 / 3

h = R - R cosθ

= R ( 1 - 2 / 3 )

= R / 3 .

he kicked it 10 feet straight up into the air? i think this works.