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The bounce of a bungee jump

  1. Apr 25, 2017 #1
    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?

    2. Relevant 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?
     
  2. jcsd
  3. Apr 25, 2017 #2

    PeroK

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    What happens if literally no energy is lost?
     
  4. Apr 25, 2017 #3
    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
     
  5. Apr 25, 2017 #4

    PeroK

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    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.
     
  6. Apr 25, 2017 #5
    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?
     
  7. Apr 25, 2017 #6

    PeroK

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    Perhaps better is that the initial gravitational PE is converted to elastic PE in the rope, then back to gravitational PE again ...
     
  8. Apr 25, 2017 #7
    Oh, thanks for your time and help!
     
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