How high will UT bounce on a bungee jump from the Golden Gate Bridge?

AI Thread Summary
UT will bungee jump from the Golden Gate Bridge at a height of 65 meters, with the rope pulling her up 8 meters above the water. Her weight is 63 kg, leading to a potential energy calculation of 35,191 J for the fall. The velocity just before the jump is calculated to be 33.45 m/s, and the distance the rope pulls her upwards is determined to be 55.4 meters. The discussion highlights that if no energy is lost, the energy from the fall is converted to elastic potential energy in the rope, allowing her to bounce back to her starting height. The conversation emphasizes the importance of understanding energy conservation in this physics scenario.
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1. Ut will bungee jump from the golden gate bridge. The height that she will jump from is 65 meters. The rope pulls her up 8 meters above the water. If UT's weight is 63 kg, how far will the rubber rope pull her upwards, if there's no loss to friction or air resistance?

Homework Equations


PE= mgh
v=√2gh
v^2-u^2=2as

3. The Attempt at a Solution

I started by calculating the energy of the fall.
PE= 63*(65-8)*9.8= 35191 J

I then tried to calculate the velocity that she reaches when she is about to fall.
v=√2gh= 33.45 m/s

I then used v^2-u^2=2as, u^2 at the top is 0 and a= g=9.8 and s=h
v^2-u^2=2as
33.45^2-0= 2*9.82* s
s= 55.4 meters

That seems oddly wrong that she could reach such a height, is this the right way to do it or is there a mistake?
 
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What happens if literally no energy is lost?
 
Well, the energy produced will be enough to bounce her back where she started, if nothing of it is lost; though I don't know how to present it with formulas to show that my reasoning is enough
 
Superman123 said:
Well, the energy produced will be enough to bounce her back where she started, if nothing of it is lost; though I don't know how to present it with formulas to show that my reasoning is enough

I would say you don't need any formulas, but you perhaps need to describe the scenario in physics terms. After all, if you push a vehicle along a flat, frictionless track, it doesn't come back to where it started.
 
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Is saying " The energy of the fall will be preserved in the rope and because there's no energy loss, the rope will pull up UT with the same energy, getting here to the same she started in." correct, considering physics terms?
 
Superman123 said:
Is saying " The energy of the fall will be preserved in the rope and because there's no energy loss, the rope will pull up UT with the same energy, getting here to the same she started in." correct, considering physics terms?

Perhaps better is that the initial gravitational PE is converted to elastic PE in the rope, then back to gravitational PE again ...
 
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Oh, thanks for your time and help!
 

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