Question

Bungee Jumping Originating on Pentecost Island in the Pacific, the practice of a person jumping from a high place harnessed to a flexible attachment was introduced to western culture in 1979 by the Oxford University Dangerous Sport Club. One important parameter to know before attempting a bungee jump is the amount the cord will stretch at the bottom of the fall. The stiffness of the cord is related to the amount of stretch by the equation

$K=\frac{2W(S+L)}{S^2}$

where W= weight of the jumper (pounds)

K= cord's stiffness (pounds per foot)

L= free length of the cord (feet)

S=stretch (feet)

(a) A 150-pound person plans to jump off a ledge attached to a cord of length 42 feet. If the stiffness of the cord is no less than 16 pounds per foot, how much will the cord stretch?

(b) If safety requirements will not permit the jumper to get any closer than 3 feet to the ground, what is the minimum height required for the ledge in part (a)?

Short answer

1. The stretch of the cord is approximately 39 ft.
2. The height of the ledge should be at least 84 ft for the 150 lb. jumper.

Step-by-step solution

Given information

Based on the given parameters in the problem, $W=150,K=16,L=42$.

Determine the value of S

Substitute $W=150,K=16,L=42$in$K=\frac{2W(S+L)}{S^2}$

role="math" localid="1649312260564" $$\begin{gathered} 16=\frac{2\left(150\right)(S+42)}{S^2} \\ 8S^2=\left(150\right)(S+42) \\ 4{\mathrm{S}}^2-75\mathrm{S}-3150=0 \end{gathered}$$

Use the quadratic formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$$\begin{gathered} S=\frac{-\left(-75\right)\pm \sqrt{{\left(-75\right)}^2-4\left(4\right)\left(-3150\right)}}{2\left(4\right)} \\ =39,-20 \end{gathered}$$

Determine the value of H.

Let H be the height of the ledge.

Since the maximum length that the cord can reach should be 3 ft above the ground, it follows:

$S+L=H-3$

Substitute $S=H-L-3$into the equation:

$K=\frac{2W(H-L-3+L)}{(H-L-3{)}^2}$

Substitute $K=16,W=150$and $L=42$ to the equation:

$$\begin{gathered} 16=\frac{2·150·(H-42-3+42)}{(H-42-3{)}^2} \\ \frac{8}{150}=\frac{(H-3)}{{(H-45)}^2} \\ 8{(H-45)}^2=180(H-3) \\ 4{\mathrm{H}}^2-435\mathrm{H}+8325=0 \end{gathered}$$

$\begin{array}{rcl}H& =& \frac{-(-435)\pm \sqrt{(-435{)}^2-4\times 4\times 8325}}{2\times 4}\\ & =& 84,25\end{array}$

Write the conclusion

1. Since the value of S is a measure of distance, the negative solution for S can be ignored. Therefore, $S=39$.
2. Since the length of the cord is $L=42\mathrm{ft}$, the height of the ledge cannot be $25\mathrm{ft}$ because it is smaller than the cord length. Therefore,$H=84ft$