A daredevil plans to bungee jump from a balloon 53.0 m above a carnival midway.

Question

A daredevil plans to bungee jump from a balloon 53.0 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 1.65 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

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