Answer to Question #127335 in Quantum Mechanics for melanie

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)

$$L+x=62.5-13=49.5$$

$$0.5kx^2=mgh\\0.5k(49.5-L)^2=mg(49.5)\\mg(5)=k(1.45)L\\$$

$$0.5(5)(49.5-L)^2=(49.5)(1.45)L\\L=23.5\ m$$

b)

$$ma=kx-mg\\ma=5mg\frac{(49.5-L)}{1.4L}-mg\\a=5(9.8)\frac{(49.5-23.5)}{1.4(23.5)}-9.8=28.9\frac{m}{s^2}$$