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Homework Help: How to find the Max. distance of the bungee jumper?

  1. Jan 16, 2005 #1
    Please help me to solve this question because I don't have a clue to do that!!

    Q: In a bungee jump a volunteer of mass 70 kg drops from a bridge, tethered to his jump point by an elastic cable of unstretched length L = 20m and elastic modulus 3000 N. Ifnoring energy losses, and assuming he hits nothing below, find the jumper's maximum distance of fall.

    Thanks!
     
  2. jcsd
  3. Jan 16, 2005 #2

    dextercioby

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    HINT:Use the law of conservation of mechanical energy.Pay attention with the 3 various types of energy the system has and with the "zero" for gravitational potential energy.

    Daniel.

    P.S.The problem is posted twice... :grumpy:
     
  4. Jan 16, 2005 #3
    The total mechanical energy before the jump is equal to the total mechanical energy after the jump.

    The relevant equation is [itex]K_1 + U_{grav,1} = K_2 + U_{grav,2} + U_{el,2}[/itex].

    [itex]\frac{1}{2}mv_1^{2} + mgy_1 = \frac{1}{2}mv_2^{2} + mgy_2 + \frac{1}{2}ky_2^{2}[/itex]

    Choosing the relaxed hanging length of the rope as the origin,

    [itex]0 + mgL = mgy_2 + \frac{1}{2}ky_2^{2}[/itex].

    Exercise for the reader: find an equation relating the spring constant to the elastic modulus of the bungee cord and solve the above equation for [itex]y_2[/itex]. Hint: the distance of fall is not [itex]y_2[/itex].
     
  5. Jan 16, 2005 #4
    Potential energy for spring constant F=- kx, k=λ*A*x/L λ; however, now (A)cross-section area of the cord is provided; for string constant F = -λ (x/L). But will they equal?! kx = λ (x/L)?!

    Sorry, still not very understand!! :frown:
     
    Last edited: Jan 16, 2005
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