The max. distance of fall of the bungee jumper?

[Esta]

Please help me to solve this question because I don't have a clue to do that!

Q: In a bungee jump a volunteer of mass 70 kg drops from a bridge, tethered to his jump point by an elastic cable of unstretched length L = 20m and elastic modulus 3000 N. Ifnoring energy losses, and assuming he hits nothing below, find the jumper's maximum distance of fall.

Thanks!

 

[enigma]

Esta,

Homework questions should be posted in the homework help section, and you should post your work, even if it's only as much as looking up relevant equations in the book.

Moving this thread to homework help

-enigma
-Engineering Mentor

 

[wais]

A: To solve this question, we can use the equation for potential energy: PE = mgh, where m is the mass of the jumper, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the jumper above the ground.

First, we need to find the height that the jumper will fall from. Since the unstretched length of the bungee cord is 20m, this means that the jumper will fall a maximum of 20m before the bungee cord starts to stretch.

Next, we need to find the potential energy at the top of the fall. We know that the jumper's mass is 70kg, so we can plug this into the equation: PE = (70kg)(9.8 m/s^2)(20m) = 13,720 J.

Now, we need to find the potential energy at the bottom of the fall. Since the bungee cord will stretch as the jumper falls, we need to take into account the elastic potential energy of the cord. We can use the equation for elastic potential energy: PE = (1/2)kx^2, where k is the elastic modulus (3000 N) and x is the distance the cord stretches.

To find x, we can use Hooke's Law: F = kx, where F is the force on the cord. At the bottom of the fall, the force on the cord will be equal to the weight of the jumper, so we can set the equations equal to each other: (70kg)(9.8 m/s^2) = (3000 N)x. Solving for x, we get x = 2.27 m.

Plugging this value into the equation for elastic potential energy, we get PE = (1/2)(3000 N)(2.27m)^2 = 7,690 J.

Now, we can find the maximum distance of fall by subtracting the potential energy at the bottom of the fall from the potential energy at the top of the fall: 13,720 J - 7,690 J = 6,030 J.

Therefore, the maximum distance of fall for the bungee jumper will be 6,030 J.