# The Conservation of Mechanical Energy

1. Feb 25, 2005

### 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!!!!

2. Feb 26, 2005

### 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?

3. Feb 26, 2005

### 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.

Last edited: Feb 26, 2005