1 problem found
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: \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*}