Energy air resistance of a sky diver

Question

An 85 kg skydiver is falling through the air at a constant speed of 195 km h^-1. At what rate does air resistance remove energy from the skydiver?

Answer 1

To solve this, we start with the drag force from air resistance. For high velocity, we can approximate this as F = (1/2)ρCAv². Here A is the cross-sectional area of the object, ρ is the density of the fluid (air in this case), C is a drag coefficient and the velocity is v. As a dissipative force, the total energy dissipated over a distance x will be F*x, where F is the component of the force that points in the direction of x. The rate of change of the energy dissipated will then be P = F*v for a constant force, or P = (1/2)ρCAv³.

However, we only know the mass of the skydiver, not their area! Well, one trick is that if the diver is falling at a constant rate, they must be at terminal velocity, their maximum possible velocity. This is where the frictional force balances the gravitational force, such that there is no acceleration. We thus must have that (1/2)ρCAv² = mg, where the mass is m and g is the gravitational constant. But then the rate of energy dissipation is nothing more than mgv.

For our case, this is (85 kg)*(195 km/hr)*(9.81 m/s²). Applying the appropriate conversion from hours to seconds and from kilometers to meters, this is 45 kilowatts.