Conservation of energy with a bungee jumper

1. Starting from rest, a 64kg person bungee jumps from a tethered balloon 65m high. The cord has unstretched length 25.8m. The cord is modeled as a spring that obeys hooke’s law with spring constant 81N/m and the person’s body is modeled as a particle. The balloon does not move. a) Express the gravitational potential energy of the person as a function of the person’s variable height y above the ground. Express the elastic potential energy of the cord as a function of y. Assume air resistance is negligible, determine the minimum height of the person above the ground during his plunge.

2. Relevant equations
Hooke’s law f = -kx
Work done by a spring = (1/2)kx^2
Gravitational potential energy U =mgy

3. I have no idea where to start here. I used mgy for the first one. Second one I put (1/2)(81)(25.8-y)^2. For the final one I used mgy to find the person’s potential energy at the very top (40,768J)before the jump and used that as total energy. Then, I equated that to mgy + 1/2(81)(25.8-y)^2 and solved which gave me y as 1.16m. I don’t think that’s correct though since a bungee jumper would risk going that close to the ground before the cord pulls him back up.

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