Chapter 4.6, Problem 79AYU

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

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

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)?

Source: American Institute of Physics, Physics News Update, No. 150, November 5,1993.

To determine

To find:

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?

Answer to Problem 79AYU

a. 39 ft

Explanation of Solution

Given:

A 150-pound person plans to jump off a ledge attached to a cord of length 42 feet.

Calculation:

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

where $W$ weight of the jumper (pounds), $K$ cord’s stiffness (pounds per foot), $L$ free length of the cord (feet), $S$ stretch (feet)

$16 \le \frac{2\left(150\right)\left(S + 42\right)}{S^2}$

$16S^2 - 300S - 12600 \le 0$

The zero of $16S^2 - 300S - 12600 = 0$

$S = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$S = \frac{300 \pm \sqrt{{\left(-300\right)}^2 - 4\left(16\right)\left(-12600\right)}}{2\left(16\right)}$

$S = \frac{300 \pm \sqrt{90000 + 64\left(12600\right)}}{32}$

$S = \frac{300 \pm \sqrt{896400}}{32}$

$S = 38.96, -20.21$

The cord stretch is approximately not less than 39 pounds.

To determine

To find:

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)?

Answer to Problem 79AYU

b. 84 ft

Explanation of Solution

Given:

A 150-pound person plans to jump off a ledge attached to a cord of length 42 feet.

Calculation:

b. spring potential energy.

$\text{PS = ½kx²}$

potential energy of position.

$\text{PE = mgh}$

where

$m$ is mass of jumper.

$k$ is spring constant of bungee.

$x$ is length of bungee stretch.

$h$ is total height of fall.

$mgh = \mathrm{½}kx²$

$150h = \frac{1}{2}\left(16\right)\left(h - 42\right)²$

$300h = 16\left(h^2 + 1764 - 84h\right)$

$16h^2 + 28224 - 1344h - 300h = 0$

$16h^2 - 1644h + 28224 = 0$

$h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$h = \frac{1644 \pm \sqrt{{\left(1644\right)}^2 - 4\left(16\right)\left(28224\right)}}{32}$

$h = 80.96,21.79$

$h \approx 81$

$\text{platform height = height of drop plus safety margin = 81 + 3 = 84 ft}$.

Therefore the ledge should be at least 84 ft above the ground for a 150 pound jumber.