Question

A daredevil wishes to bungee-jump from a hot-air balloon 62.5 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 13.0 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 1.45 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?
m

(b) What maximum acceleration will he experience?
m/s²

Answer 1

a) The daredevil bunjee jumps from the hot air balloon at a height h1 above the carnival highway where h1=62.5 m

The daredevil stops at a height h2 from the ground where h2=13 m

Thus, the distance of his travel h=62.5-13=49.5 m

In the preliminary test, the weight of the body stretches the 5 m cord by 1.45 m

Thus, Stretch of the cord per metre=1.45/5=0.29 m

Thus x+0.29x=49.5

where x= length of the cord initially

Thus x=38.37 m

b) If there is a retardation due to some friction/air drag, etc, the effective acceleeration can be written as (g-a) m/s²

In the absence of such retarding force, a=0, which gives the maximum acceleration g=9.8 m/s²