Question

Unstretched length of the bungee cord is L=15 m. Bungee-jumper has a mass of 60 kg. Stiffness of the cord is k=300 N/m. Determine the lowest point that the jumper would reach, assuming that his/her initial velocity is zero. The weight of the cord is neglected.Center of gravity of the jumper is 1.0 m. above the ground

Answer 1

Since the system is conservative (only potential energies of the cord and the jumper are involved) we can write law of conservation of energy using

$$h$$

- height of the center of gravity above the ground,

$$x$$

- the elongation of the spring, and assuming that

$$g=10$$

:

$$mg(L+h+x)=\frac{kx^2}{2},$$ $$x^2 \cdot \frac{k}{2}-x\cdot mg-mg(h+L)=0,$$ $$x^2 \cdot \frac{300}{2}-x\cdot 60\cdot 10-60\cdot 10(1+15)=0,$$

solve this equation and choose the correct root:

$$x=10.24 \text { m}.$$

Thus the lowest point is

$$P=L+x=15+10.24=25.24 \text{ m}$$

below the surface of the bridge.