# Bungee Jumping Question

1. Apr 8, 2014

### patrickmoloney

1. The problem statement, all variables and given/known data

A bungee jumper of mass m drops off a bridge and falls vertically downwards.
The bungee cord is elastic with natural length L and stiffness k. Deduce that
at the lowest point of the fall, the cord is stretched by an amount $$x = \frac {mg}{k} \Big( 1 + \sqrt {1+ \frac {2kL}{mg}} \Big)$$

2. Relevant equations

$mgh = \frac{1}{2} k x^2$

3. The attempt at a solution

Energy considerations dictate that the gravitational potential energy of the jumper in the initial state is equal to the elastic potential of the cord in the final state

$mg ( L + x ) = \frac{1}{2} k x^2$

$mgL + mgx = \frac{1}{2} k x^2$

$2mgL + 2mgx = k x^2$

$x^2 - \Big( \frac{2mg}{k} \Big )x - \Big( \frac{2mgL}{k} \Big) = 0$

$x = \frac{- \Big(- \frac{2mg}{k} \Big) + \sqrt{ \Big( - \frac{2mg}{k} \Big)^2 - 4 \Big( - \frac{2mgL}{k} \Big )}}{2}$

$x = \frac{1}{2} \Big( \frac{2mg}{k} + \sqrt{ \frac{4m^2g^2}{k^2} + \frac{8mgL}{k}} \Big)$

$x = \frac{mg}{k} + \frac{1}{2} \Big ( \sqrt{ \frac{4m^2g^2}{k^2} \Big ( 1 + \frac{2kL}{mg} \Big )} \Big )$

$x = \frac{mg}{k} + \frac{1}{2} \Big( \sqrt{\frac{4m^2g^2}{k^2}} \sqrt{1+\frac{2kL}{mg}} \Big )$

$x = \frac{mg}{k} + \frac{1}{2} \Big( \frac{2mg}{k} \Big) \sqrt{1 + \frac{2kL}{mg}}$

$x = \frac{mg}{k} + \frac{mg}{k} \sqrt{1 + \frac{2kL}{mg}}$

$x = \frac{mg}{k} \Big( 1 + \sqrt{1+ \frac{2kL}{mg}} \Big)$

Is this the correct method? There is no solution online to this problem.

2. Apr 8, 2014

### CWatters

Seems reasonable to me.