physics
posted by sharon on .
an extreme skier starting from rest, coasts down a mountain that makes an angle 25.0¨¬with the horizontal. The coeffiecient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 8.4m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 2.80m below the edge. How fast is she going just before she lands?

Frictional force, F
= μmg cos(θ)
Work done to overcome kinetic friction
= F*D
= 8.4F
Difference in elevation, H
= 8.4 sin(&theta)+2.8 m
Energy left (in the form of KE)
= mgH8.4F
Equate Kinetic energy with potential energy
(1/2)mv² = mgH8.4(μmg cos(θ))
All quantities are known except v.
Solve for v (9.7 m/s)
Check my thinking and calculations.