A daredevil plans to bungee jump from a balloon 72.0m above the ground. He will

Question

A daredevil plans to bungee jump from a balloon 72.0m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at point 8.00m above the ground. Model his body as a particle and the cord as having negligible mass and obeying Hookes 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 1.35 m. he will drop from rest at 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?

Explanation / Answer

part(a)

Let the length of the cord is L .

m= mass of jumper

according to Hook's law

F = k*x

k = F/x,where F = m*g       x = 1.35m

For the cord, k = m*g/1.35,

When jumper falls through the height L, he loses Potential energy   and the cord becomes taut. afterthat, it starts stretching and stretches till the PE lost by the jumper and converted in to kinetic energy .
Let the cord stretch by x' meter

m*g*L = (1/2)*k*x^2

and L+x' = 72 - 8

L + x' = 64

x' = 64-L

m*g*L = (1/2)*(m*g/1.35)*((64-L)^2)

2.70*L = L^2 - 128*L + 4096

L^2 - 130.70*L + 4096 = 0

L = 78.56 or 52.13

length of cord should be 52.13 m.

part(b)

restoring force in the cord in the upward direction

= k*x'      (x' id elongation in the cord)
= (mg/1.35)*( 64 - 52.13 )
= mg * (11.87/1.35)

Downward force = m*g

Net force upwards = m*g*(11.87/1.35) - mg = mg(10.52/1.35)

Net force = m*a

Maxm. acceleration upwards

a = Net force/m

a = mg(10.52 / 1.35)/m

a = 76.44 m/s^2

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