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

1. Jan 16, 2005

Esta

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. Jan 16, 2005

dextercioby

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:

3. Jan 16, 2005

jdstokes

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 $y_2$.

4. Jan 16, 2005

Esta

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!!

Last edited: Jan 16, 2005