Question

It’s your birthday, and to celebrate you’re going to make your first bungee jump. You stand on a bridge $100m$above a raging river and attach a $30m$-long bungee cord to your harness. A bungee cord, for practical purposes, is just a long spring, and this cord has a spring constant of $40N/m$. Assume that your mass is $80kg$. After a long hesitation, you dive off the bridge. How far are you above the water when the cord reaches its maximum elongation?

Short answer

The distance between you and water when the bungee cord is at maximum elongation$=10.9m$.

Step-by-step solution

Given information

Height of the bridge,$h=100m$

Length of bungee cord $L=30m$

The spring constant of the cord$k=40N/m$

Explanation

The potential energy of the cord:

$$\begin{gathered} \text{Potentialenergy}=mgh \\ wherem=80kg,g=9.8m/s^2 \\ \text{Potentialenergy}=\left(80kg\right)\left(9.8m/s^2\right)h \\ =784hJ \end{gathered}$$

The elastic energy of the cord,

role="math" localid="1649402893356" $$\begin{gathered} U_x=\frac{1}{2}k{y}^2 \\ =\frac{1}{2}\left(40N/m\right)y^2 \\ =\left(20N/m\right)y^2 \end{gathered}$$

role="math" localid="1649402871543" $$\begin{gathered} \text{Distance}=h+y \\ =\left(30m\right)+y \end{gathered}$$

By conservation of energy, equate potential energy and elastic energy.

role="math" localid="1649403050726" $$\begin{gathered} \left(20N/m\right)y^2=\left(784kg.m/s^2\right)h \\ h=\left(100m\right)+y \\ \left(20N/m\right)y^2=\left(784kg.m/s^2\right)\left(100+y\right) \\ 20y^2=784y+78400 \\ 20y^2-784y-78400=0\cdots \cdots \cdots \left(1\right) \\ \text{Bysolvingtheequation}\left(1\right) \\ y=85.21m \end{gathered}$$

Distance of the cord$$\begin{gathered} =\left(100m\right)-\left(30m+85.21m\right) \\ =-15.21m \end{gathered}$$

The distance of the cord is $15.21m$below the water.