A daredevil plans to bungee jump from a balloon 70.0 m above a carnival midway.
ID: 1337620 • Letter: A
Question
A daredevil plans to bungee jump from a balloon 70.0 m above a carnival midway. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 10.0 m 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, hanging at rest from a 5.00 m length of the cord, he finds that his body weight stretches it by 1.20 m. He will drop from rest at the 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
PART A
Initially lenght of the cord is = 5 m
The extension is 1.65 m
k = mg/x = mg/1.65
k' = kl/L = mg/1.65*5/L
yi = 53 m
yf = 10 m
xf = (yi - yf ) - L
mg (yi - yf ) = 1/2* k'*xf2
mg ( 53 - 10) = 1/2*((mg/1.65)* (5/L)*(43-L)2
43 = 1/2*(3.03/L)(43-L)2
L = 19.48 m
PART B
xmax = x = 43 - L
k'xmax - mg = ma
(mg/1.65)(5/L)*(43 - L) - mg = ma
(g/1.65)(5/L)(43-L) - g = a
a = (9.8/(1.65)(5/19.48)(43-19.48) - 9.8
a = 26.06