A daredevil plans to bungee jump from a balloon 65.0 m above the ground. He will

Question

A daredevil plans to bungee jump from a balloon 65.0 m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his a point 10.0 m fall at 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 5.0 m length of the cord, his body weight stretches it by 1.5 m. He will drop from rest at the point where the top end of a longer section of the cord attached to the stationary balloon. a) what length of cord should he use? b) What maximum acceleration will he experience?

Explanation / Answer

h = 65 m ; h' = 10 m ; l = 5 m ; x = 1.5 m ;

a)We need to know the soring constant first,

for a vertical spring, mg = kx

k = mg/x = m x 9.8/1.5 = 6.53 m

PE at h = (65 - 10) = 55 m

PE = m g h = m 9.8 x 55 = 539 m

from conservation of energy

1/2 k x^2 = m g h = 539 m

1/2 x 6.53 m x^2 = 539 m

x = 12.85 m

So the length of the cord will be:

L = 55 - 12.85 = 42.15 m

Hence, L = 42.15 m

b)The max acceleration would be at the bottom, when the speing is fulley streched so

a = kx/m - g = 6.53 m x 12.85/m - 9.8 = 74.11 m/s^2

Hence, a = 74.11 m/s^2

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