# Chapter 10 - Energy - Exercises and Problems - Page 276: 70

When the bungee reaches maximum elongation, the person is 10.9 meters above the water.

#### Work Step by Step

We can use conservation of energy to solve this question. The sum of the energy stored in the bungee and the final potential energy will be equal to the initial potential energy. Note that the bungee begins to stretch after the person falls 30 meters Let $x$ be the distance that the bungee stretches. $U_s+PE_f = PE_0$ $\frac{1}{2}kx^2+mg(100-30-x) = mg~(100)$ $\frac{1}{2}kx^2+ mg(-x-30) = 0$ $\frac{1}{2}(40~N/m)x^2+ (80~kg)(9.80~m/s^2)(-x-30) = 0$ $(20~N/m)~x^2-(784~N)~x-23,520~J = 0$ We can use the quadratic formula to find $x$. $x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $x = \frac{-(-784)\pm \sqrt{(-784)^2-(4)(20)(-23,520)}}{(2)(20)}$ $x = -19.9~m, 59.1~m$ Since the negative value is unphysical, the solution is $x = 59.1~m$. The height above the water is $100~m-30~m-59.1~m$ which is 10.9 meters. When the bungee reaches maximum elongation, the person is 10.9 meters above the water.

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