Answer
10.1 m
Work Step by Step
Let's apply the principle of conservation of mechanical energy to the system.
Kinetic energy + Potential energy = Constant
$mgH_{I}=mgH_{F}+\frac{1}{2}kx^{2}-(1)$
According to the given figure, we can write
$x=H_{F}-H_{A}-(2)$ ; Where $H_{A}=37\space m$
(2)=>(1),
$mgH_{I}=mgH_{F}+\frac{1}{2}k(H_{F}-H_{A})^{2}$
$2mgH_{I}=2mgH_{F}+k(H_{F}^{2}-2H_{F}H_{A}+H_{A}^{2})$
$kH_{F}^{2}+2(mg-kH_{A})H_{F}+(kH_{A}^{2}-2mgH_{I})=0$ ; Let's plug known values into this equation.
$(66\space N/m)H_{F}^{2}+2(68\space kg\times9.8\space m/s^{2}-66\space N/m\times37\space m)H_{F}+[66\space N/m(37\space m)^{2}-2(68\space kg)(9.8\space m/s)(46\space m)]=0$
$(66\space N/m)H_{F}^{2}-(3551.2\space N)H_{F}+29045.2\space Nm=0$
This is a quadratic equation, so we can use the quadratic formula to find the value of $H_{F}$
$H_{F}=\frac{-(-3551.2\space N)\space \pm \sqrt {(-3551.2\space N)^{2}-4(66\space N/m)(29045.2\space Nm)}}{2\times66\space N/m}$
$H_{F}=45.2\space m \space or \space 10.1\space m$
The value 45.2 m is above the point where the bungee cord is stretched. So we can neglect that solution. Therefore, at the lowest point height above the water = 10.1 m
