Question

Multiple-Concept Example 5 reviews many of the concepts that play a role in this problem. An extreme skier, starting from rest, coasts down a mountain that makes an angle of 26.9 ° with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.213. She coasts for a distance of 19.9 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 4.92 m below the edge. How fast is she going just before she lands?

Answer 1

here,

the angle of inclination , theta = 26.9 degree

uk = 0.213

the distance traveled , s = 19.9 m

the acceleration of skier , a = (m * g * sin(theta) - uk * m * g * cos(theta)) /m

a = 9.81 * (sin(26.9) - 0.213 * cos(26.9)) m/s^2

a = 2.57 m/s^2

the velocity of skier before leaving the ramp , u = sqrt(2 * a * s)

u = sqrt(2 * 2.57 * 19.9) m/s

u = 10.1 m/s

h = 4.92 m

let the speed of skier just before it lands be v

using conservation of energy

0.5 * m * v^2 = 0.5 * m * u^2 + m * g * h

v^2 = 10.1^2 + 2 * 9.81 * 4.92

solving for v

v = 14.1 m/s

the speed of skier just before it lands is 14.1 m/s