A skydiver of mass m jumps from a hot air balloon and falls a distance d before reaching a terminal velocity of magnitude v. Assume that the magnitude of the acceleration due to gravity is g.-What is the work (Wd) done on the skydiver, over the distance , by the drag force of the air?
-Find the power (P d) supplied by the drag force after the skydiver has reached terminal velocity v.

Using the work energy theorem, the work done by the drag force while the skydiver moves through a distance d is equal to the change in kinetic energy and potential energy of the body.

The potential energy of the skydiver drops by a magnitude of (mgd) falling through a distance of d.

ΔP.E = - mgd (skydiver loses this amount of potential energy)

But the skydiver falls from rest to gain a velocity of v after falling a distance of d vertically.

ΔK.E = (final kinetic energy at distance d) - (initial kinetic energy at rest)

Final kinetic energy at d = (1/2)(m)(v²) (the body is now falling with terminal velocity (v) at this point d)

Initial kinetic energy = 0 (since the skydiver was initially at rest)

ΔK.E = (1/2)(m)(v²) - 0 = (mv²/2)

W = ΔK.E + ΔP.E = (mv²/2) - mgd

b) At terminal velocity, the net force on the skydiver = 0

hence,

drag force = weight of the skydiver = mg

Power = F × v = mg × v = mgv

Hope this Helps!!!