# Bungee Jump: Conservation of Energy

1. May 19, 2014

### teme92

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

A bungee jumper of mass m drops o ff a bridge and falls vertically downwards. The bungee cord is elastic with natural length L and stiff ness k. Deduce that at the lowest point of the fall, the cord is stretched by an amount:

x=mg/k(1+√(1+(2kl/mg))

2. Relevant equations

F=-kx
PE=mgh=0.5kx2

Where h=L+x

3. The attempt at a solution

The total energy before the jump is equal to the total energy after the jump. Since at the bottom there is no kinetic energy I said the Potential Energy before is equal to the Elastic Energy at the bottom.

So:

mg(L+x)=0.5kx2
x2=(2mgx +2mgl)/k
x=√((2mgx +2mgl)/k)

I'm having problems here trying to get in the asked form. Have I forgotten something n the conservation of energy?

2. May 19, 2014

### willem2

The line above that is a quadratic equation.

3. May 19, 2014

### teme92

So does that mean I have to solve it quadratically?

4. May 19, 2014

### willem2

I suppose you know how to solve a quadratic equation. ax^2 + bx + c = 0 ?