A bungee jumper of mass 60kg jumps from a bridge 24 m above the surface of the water. The rope is 12 m long and is assumed to obey Hooke's law. What should the spring constant of the rope be if the woman is to just reach the water?
The Attempt at a Solution
Energy conservation? I'm definitely doing something wrong here and not understanding something but I don't know what.
mgh = ½kx^2
(60)(10)(12) = ½k(12)^2
[I put the length of the unstretched part of the rope as my h value and x as the extension, which is 24-12=12]
k= 100 Nm^-1
[This is the answer I got]
But the answer is k=200 Nm^-2