A girl starting from rest at the top of a 40 m jump, inclined at 30º above the horizontal, has 0.15 coefficient of friction. The jump ends with a frictionless, horizontal part. How far does she land away from the jump, which is 50 m above the horizontal land below?
height of slide above release point of slide = 40
friction force parallel to slide = .15 * m g cos 30
work done by friction = (40/sin 30) m g cos 30 = 20 m g cos 30
so at release from slide:
(1/2) m v^2 = 20 m g - 20 m g cos 30
v^2 = 40 g (.134) = 52.6
v = 7.25 m/s horizontal
so
d = 7.25 t
what is t, time to fall 50 m?
50 = 4.9 t^2
t = 3.19 seconds
so
d = 7.25 * 3.19 = 23.2 meters