# How to find the Max. distance of the bungee jumper?

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!

dextercioby
Homework Helper
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:

Esta said:

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!

The total mechanical energy before the jump is equal to the total mechanical energy after the jump.

The relevant equation is $K_1 + U_{grav,1} = K_2 + U_{grav,2} + U_{el,2}$.

$\frac{1}{2}mv_1^{2} + mgy_1 = \frac{1}{2}mv_2^{2} + mgy_2 + \frac{1}{2}ky_2^{2}$

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

$0 + mgL = mgy_2 + \frac{1}{2}ky_2^{2}$.

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 $y_2$. Hint: the distance of fall is not $y_2$.

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!!

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