A daredevil plans to bungee jump from a balloon 65.0m above the ground. He will
ID: 2302097 • Letter: A
Question
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 Hookes 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?
Explanation / Answer
a. For the cord, k = mg/1.5, where m = mass of the jumper.
Let L = length of the cord required.
When he falls through the height L, he loses PE = mgL and the cord becomes taut. Thereafter, it starts stretching and stretches till the PE lost by the jumper is stored in the cord in the form of PE of the cord.
Let the cord stretch by x meter
=> mgL = (1/2)kx^2 and L+x = 65 - 10 = 55
=> mgL = (1/2)*(mg/1.5)*(55-L)^2
=> 3L = L^2 - 110L + 3025
=> L^2 - 113L + 3025 = 0
=> L = (1/2)[113