air drag = (1/2) rho v^2 A
= (1/2)(1.2)(v^2)(1.8)(.4) = 0.432 v^2
friction drag = 0.06 (70*9.81)cos 30 = 35.7 Newtons
component of weight down slope = (70*9.81) sin 30 = 343 Newtons
so
343 = 35.7 + .432 v^2
solve for v
What is the terminal velocity for a 70 kg skier going down with a 30 degree snow covered hill, on wooden skis. Skier is 1.8 m tall and 0.4 wide. We have a friction of 0.06, C = 1.1, and P = 1.2 kg/m^3
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