Question

A bungee jumper of mass 75kg has a bungee cord of length L=35m. The spring constant of the cord is 92.5N/m. 1. From how high can she safely jump? 2. What is her maximum speed? 3. After many oscillations, air resistance eventually slows her down until she stops. Is her final distance from her starting position less than, greater than, or equal to 35m? Explain your reasoning.

Answer 1

To solve the equation we proceed to use the law of conservation of the material, so

For A)

`mgh = (1)/(2)k(h-L)^2`

`75*9.8*h = (92.5)/(2)(h-35)^2`

Solving for h,

`h= 67.8m`

For B)

First obtain x,

`mg=kx`

`x= (mg)/(k) = (75*9.8)/(92.5) = 7.95m`

Then,

`mg(L+x) - (1)/(2)kx^2 = (1)/(2)mv^2`

`(75)(9.8)(42.95)-(1)/(2)(92.5)(7.95)^2 = (1)/(2)(75)v^2`

`v= 27.6 m/s`

For 3)

Greater than. Because the final distance is greather than 35mm.