The Conservation of Mechanical Energy

[zizikaboo]

help I'm really stuck on this question!
A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill. The crest of the second hill is circular, with a radius of r = 45 m. Neglect friction and air resistance. What must be the height h of the first hill so that the skier just loses contact with the snow at the crest of the second hill?

um. i know it has to do with 1/2m(vf)^2+mghf=1/2mv0^2+mgh0 but i just don't really understand what the question meant by 2nd hill and its radius!

 

[Davorak]

Think of the centripetal acceleration as he is at the top of the second hill?
What does the acceleration need to be if he barely going to lift off?

 

[Curious3141]

By conservation of energy :

PE at start (crest of first hill) = (PE + KE) at end (crest of second hill).

Also, while moving in a circular trajectory on the second hill,

Weight of skier - Normal reaction force = Centripetal force.

What happens to the reaction force when the guy just loses contact ?

What's the expression for the centripetal force in terms of mass, velocity and radius ? How is it related to the expression for kinetic energy ?

Now plug in the expressions for each of those and solve for the height in terms of the radius of curvature of the crest of the second hill.