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.