Problem source

A bungee-jumper of mass $m$ is attached by means of a light rope of natural length $l$ and modulus of elasticity $mg/k,$ where $k$ is a constant, to a bridge over a ravine. She jumps from the bridge and falls vertically towards the ground. If she only just avoids hitting the ground, show that the height $h$ of the bridge above the floor of the ravine satisfies 
\[
h^{2}-2hl(k+1)+l^{2}=0,
\]
and hence find $h.$ Show that the maximum speed $v$ which she attains during her fall satisfies 
\[
v^{2}=(k+2)gl.
\]

Solution source

\begin{align*}
&& \text{Energy at the top} &= mgh \\
&& \text{Energy at the bottom} &= \frac12\frac{\lambda (h-l)^2}{l}  \\
\Rightarrow && mgh & = \frac{\frac{mg}{k}(h-l)^2}{2l} \\
\Rightarrow && 2hkl &= (h-l)^2 \\
\Rightarrow && 0 &= h^2-2lh-2hlk+l^2 \\
&&0&= h^2-2hl(k+1)+l^2 \\
\Rightarrow && \frac{h}{l} &= \frac{2(k+1)\pm \sqrt{4(k+1)^2-4}}{2} \\
&&&= (k+1) \pm \sqrt{k^2+2k} \\
\Rightarrow && h &= l \left ( (k+1) \pm \sqrt{k^2+2k} \right)
\end{align*}
Since the negative root is less than $1$, she would have not fully extended the cord. Therefore $h = l \left ( (k+1) + \sqrt{k^2+2k} \right)$

Her maximum speed will be when her acceleration is $0$, ie $g = \text{force from cord}$ ie $mg = \frac{\lambda x}{l}$ or $x = \frac{mgl}{\lambda} = \frac{mglk}{mg} = kl$.

At this point by conservation of energy we will have

\begin{align*}
&& mgh &= mg(h-l-x) + \frac12 m v^2+\frac{1}{2} \frac{mgx^2}{kl} \\
\Rightarrow && mg\left ( l + kl \right) &= \frac12 m v^2 + \frac12 \frac{mgl^2k^2}{kl} \\
\Rightarrow && 2g\left ( l + kl \right) &=  v^2 +  glk \\
\Rightarrow && v^2 &= gl(2+k)
\end{align*}