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 = 35.5 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?

Answer 1

We have the formula for centripetal acceleration in circular motion:
\[a = \omega².r = \left( v² \over r \right)\]
where a is the axial acceleration needed for a body to describe the circle, omega is the radial speed (in rad/second), r is the radius of the circle described, and v is the tangential speed (along the circle).
we deduce v from a and r where a equals the gravitational acceleration
\[v = \sqrt{r.g}= \sqrt{35,5m.9,81m/s²}=18,6616m/s\]
Then we calculate the height needed, by using the formula for potential energy and kinetic energy and solve for height h.
\[E _{pot}= m.g.h = E _{kin}=\left( m.v² \over 2 \right)\]
h = v²/2.g = (18,6616m/s)²/(2*9.81m/s²) = 17,75 meter