Asked by Alfie

A 70kg sky diver falls 5000m.
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?

Answers

Answered by Damon
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 :)
Answered by Alfie
Thanks Damon!
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