A 70kg sky diver falls 5000m.

Question

a) by how much does her gpe decrease during her fall? Ep=mxgxh = 70x10x5000 = 3,500,000J
b) By how much does her kinetic energy increase during the fall? KE GAINED = GPE LOST = 3,500,oooJ
c) What would her speed be just before she hit the ground?
d) Why is your answer to (c) wrong?

Answer 1

c) (1/2) m v^2 = answer to b

d) The reason they want is that the person would have reached terminal velocity where air drag up = weight down before falling 5000 meters.
The reason they forgot about is that the diver also has the constant horizontal velocity of the airplane the whole time. I guess the person jumped from a stationary hot air balloon :)

Answer 2

Thanks Damon!