Question

An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of 25.0° with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts down a distance of 12.7 m 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 3.30 m below the edge. How fast is she going just before she lands?

Answer 1

Given , initial velocity (vi)= 0 m/s, θ = 25°

Now,

acceleration (a) = g sin25° - _k g cos25°

a = 9.8 sin25 - 0.2 x 9.8 cos25 = 2.365 m/s²

Let the final velocity be vf

distaplacement along the incline (d) = 12.7 m

Using , vf^2 = vi^2 + 2ad

V_f² = 0² + 2*(2.365)*12.7

Vf^2 = 60.071 m/s

Vf = 7.75 m/s

let the speed just before she lands be "V"

using conservation of energy

KE + PE at the edge of cliff = KE at bottom of cliff

(0.5) m V_f² + mgh = (0.5) m V²

V² = V_f² + 2 gh

V² = 7.75² + 2 x 9.8 x 3.3

V^2 = 124.7425

V = 11.169 m/s

Thus , the speed just before she lands is 11.169 m/s