Question

Assume that the skier is 1.8 m tall and 0.30 m wide. Assume the skier's drag coefficient is 0.80.

What is the terminal speed for an 85 kg skier going down a 40 ∘ snow-covered slope on wooden skis μk= 0.060?

Answer 1

First let's see what acceleration would be without air resistance:
To find the unbalanced forces on the skis, we should
look at the situation in a coordinate system where the x-axis is (downward) along the slope,
the y-axis normal to the slope. The skier's weight (85.0 kg times 9.8 m/s^2) can be
regarded as an along-slope component, 833 N sin(40), and a normal component,
833 N cos(40). The normal component gives us the frictional force on the skis,
which is (0.060)(833 N)(cos 40) = 38.28 Newtons. The along-slope component
of the gravitational force is 535.44 Newtons, so the unbalanced force along the slope
would be 497.16 N, and the acceleration would be F/m = 5.84 m/s^2.

Now we consider the air drag. At non-negligible speeds for macroscopic objects,
the usual equation is
f(drag) = -(1/2) C rho A v^2,
where rho = density of air = 1.26 kg/m^3,
A = 0.54 m^2
v^2 = 535.44 N / [ (0.5 C)(1.26 kg/m^3)(0.54 m^2) ]

use C = 0.80, I get a terminal speed of
v^2 = 535.44 N / [ (0.5 * 0.80)(1.26 kg/m^3)(0.54 m^2) ]
So, v = 44.31 m/s