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Velocity of bungy jumper, very confusing?

  1. Apr 13, 2013 #1
    1. The problem statement, all variables and given/known data
    A bungee jumper with a mass of 78kg tied to a 39 m cord jumps off a bridge from a height of 69 m. She falls to 6 .0m above the ground before the chord brings her momentarily to rest. Calculate the impulse exerted on the jumper by the cord as it stretches


    2. Relevant equations
    Ft = mvf - mvi
    P = mv
    Eg = mgh
    Ep = 1/2 x k^2
    Ek = 1/2 m v ^2


    3. The attempt at a solution
    Impulse = 0 - mvi

    To find vi:

    Ek1 + Eg1 = Eg2 + Ep2

    So here is where I'm stuck. The equation is impossible to solve because there are two unknowns (spring constant and velocity)
     
  2. jcsd
  3. Apr 13, 2013 #2

    rock.freak667

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    Why would you need the spring constant for? Once you have the velocity, you can get the change in momentum as you have the mass.
     
  4. Apr 13, 2013 #3
    I need the spring constant because some energy is converted into elastic potential energy.

    Ek1 + Eg1 = Eg2 + Ep2

    That is the equation I've made. LS = the instant before falling, RS = on stop

    So it comes out to 0.5mv^2 = mgh = mgh + 0.5kx^2

    In order to find the velocity I have to have k but its an unknown. Am I doing this right? Did I set it up correctly? I know there has to be some potential energy because the rope is only 39 m long and it stretches far beyond that point as the person falls around 69 m - 6m and the rope isnt that long
     
  5. Apr 13, 2013 #4

    TSny

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    Actually you won't need the spring constant to find the impulse of the cord. You wrote

    impulse = 0 - mvi.

    This will be the key! Think about the meaning of vi. What point of the jump does this refer to?

    Also, for the left side of the equation do you think that represents the impulse of the cord, the impulse of the force of gravity, or the total impulse of all of the forces acting on the jumper?
     
  6. Apr 13, 2013 #5

    TSny

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    Oops, You might indeed need the spring constant. But you should be able to find it.
     
  7. Apr 13, 2013 #6

    TSny

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    What is the velocity at "the instant before falling"?
     
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