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.

a) W = ΔK.E + ΔP.E = (mv²/2) - mgd

b) Power supplied by the drag force after the skydiver has reached terminal velocity = mgv

Explanation:

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!!!

answer; equivalent to a one(1) horsepower motor;