# Question #96f24

Jun 3, 2015

Have a look: Jun 4, 2015

Here's how you could approach this problem.

Think about the conservation of energy. The man starts atop of the bridge with only potential energy. At the bottom of his drop, when the cord is fully stretched, he will end up with only elastic energy.

All the change in potential energy will now be stored in the bungee cord. Assuming that the stretch in the bungee cord is equal to $x$ meters, you'd get

${E}_{\text{potential}} = m \cdot g \cdot h$, where

$h$ - the total distance of the fall.

Since you know the length of the bungee cord to be equal to 20 m, you can write $h$ as being equal to

$h = l + x = 20 + x$

At the bottom of the fall, you have

${E}_{\text{elastic}} = \frac{1}{2} \cdot K {x}^{2}$

The elastic energy depends on the stretch of the bungee, which is $x$ meters.

As a result, you'll have

${E}_{\text{potential" = E_"elastic}}$

$m \cdot g \cdot \left(l + x\right) = \frac{1}{2} K {x}^{2}$

$m g \cdot l + m g \cdot x = \frac{1}{2} K {x}^{2}$

$80 \cdot 9.81 \cdot 20 + 80 \cdot 9.81 \cdot x = \frac{1}{2} \cdot 100 \cdot {x}^{2}$

$50 {x}^{2} - 748.8 x - 15969 = 0 \implies x = \text{27.2 m}$

Thus, the total distance of the fall will be equal to

$h = 20 + x = 20 + 27.2 = \textcolor{g r e e n}{\text{47.2 m}}$

The maximum force will be exerted when the cord is at maximum stretch

$F = k \cdot x = 100 \cdot 27.2 = \textcolor{g r e e n}{\text{2720 N}}$

The maximum velocity during the fall will occur just before the bungee cord starts to stretch, i.e. after 20 m of free fall. Use the equation

$l = {V}_{0} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$, where

${v}_{0}$ - the initial velocity of the jumper, equal to zero in your case.

Solve the above equation to get the time of free fall

$20 = \frac{1}{2} \cdot g \cdot {t}^{2} \implies t = \sqrt{\frac{2 \cdot 20}{9.81}} \cong \text{2 s}$

Now use the equation

$v = {v}_{0} + g \cdot t$

$v = g \cdot t = 9.81 \cdot 2 = \textcolor{g r e e n}{\text{19.6 m/s}}$