Question

A daredevil wishes to bungee-jump from a hot-air balloon 73.0 m above a carnival midway. He will use a piece of uniform elastic cord tied to a harness around his body to stop his fall at a point 12.0 m above the ground. Model his body as a particle and the cord as having negligible mass and a tension force described by Hooke's force law. In a preliminary test, hanging at rest from a 5.00-m length of the cord, the jumper finds that his body weight stretches it by 1.85 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?

Answer 1

a)
let k is the spring constant of the cord,

use Hook's law,

F = k*x
'
==> k = F/x

= m*g/x

= m*g/1.85

let y is the extension of the cord when bungee-jumper jumps and goes to the lowest point.

apply conservation of energy

m*g*h = (1/2)*k*y^2

m*g*(73 - 12) = (1/2)*(m*g/1.85)*y^2

61 = y^2/(2*1.85)

==> y = sqrt(61*2*1.85)

= 15.0 m

so, length of the cord should be used, L = (15/1.85)*5

= 40.5 m <<<<<<<------------Answer

b) when the daredevil comes to lowest point he experiances maximum acceleration.

use, Fnet = F_cord - m*g

m*a = k*y - m*g

m*a = (m*g/1.85)*15 - m*g

a = g*15/1.85 - g

= 9.8*15/1.85 - 9.8

= 69.6 m/s^2 <<<<<<<<<<<<----------------Answer