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?
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 :)
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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