A daredevil plans to bungee jump from a balloon 65.0m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at point 10.0m above the ground. Model his body as a particle and the cord as having negligible mass and obeying Hooke❝s law. In a preliminary test he finds that when hanging at rest from a 5.00-m length of the cord, his body weight stretches it by 1.50 m. he will drop from rest at point where the top end of a longer section of the cord is attached to the stationary balloon. (a)What length of cord should he use? (b) What maximum acceleration will he experience?
Potential energy = mgh
Spring potential energy = kx^2/2
Spring force = kx
The Attempt at a Solution
I tried using conservation of energy to find the spring constant k. I said the stretched length of the short cord was h1 and the distance it stretched was x1:
mgh1 = kx1^2/ 2 => k = (2mgh1) / x1^2
I did the same with the longer cord:
mgh2 = kx2^2 / 2 => x2^2 = 2mg(h2) / k
Then I replaced k with what I found:
x2^2 = 2mg(h2) / (2mg(h1)/x1^2) => x2^2 = 2mg(h2) x1^2 / 2mg(h1) = (h2) x1^2 / h1
Is what I did right or not at all? I'm very confused about this problem, thanks for your help!