In downhill speed skiing a skier is ed by both the air drag force on the body and the kinetic frictional force on the skis. Suppose the slope angle is θ = 39.5°, the snow is dry snow with a coefficient of kinetic friction μk = 0.0400, the mass of the skier and equipment is m = 85.0 kg, the cross-sectional area of the (tucked) skier is A = 1.30 m2, the drag coefficient is C = 0.150, and the air density is 1.20 kg/m3.

Answer 1

first
Air friction drag = (1/2} rho v^2 A = 0.6*v^2*1.30 = 0.78 v^2
then do problem (If v is constant solve for v)
gravity force down slope = m g sin 39.5
forces up slope = 0.78 v^2 + 0.04*85*9.81* cos 39.5
if a = 0, down slope force = up slope force.
.78 v^2 + 0.04*85*9.81* cos 39.5 = 85*9.81*sin39.5