A daredevil plans to bungee jump from a balloon 72.0 m above the ground. He will
ID: 2135981 • Letter: A
Question
A daredevil plans to bungee jump from a balloon 72.0 m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 15.0 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 he finds that when hanging at rest from a 5.00-m length of the cord, his body weight stretches it by 1.75 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
A daredevil plans to bungee jump from a balloon 72.0 m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 15.0 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 he finds that when hanging at rest from a 5.00-m length of the cord, his body weight stretches it by 1.75 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
(a) What length of cord should he use?
m
(b) What maximum acceleration will he experience?
m/s2
A daredevil plans to bungee jump from a balloon 72.0 m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 15.0 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 he finds that when hanging at rest from a 5.00-m length of the cord, his body weight stretches it by 1.75 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
Explanation / Answer
Weight = mg
Stiffness of cord = mg/1.75
energy conservation: Potential energy change = Energy stored in cord
mg(h1 - h2) = 1/2*kx^2
mg*(72-15) = 1/2*(mg/1.75) *(72 - 15 - L)^2
Solving this, L = 42.875 m
b)
max. acceleration = g = 9.81 m/s^2