What length of rope do i need to climb the matterhorn?

Physics - Bungee problem - Find length of rope?

• Hi, My physics teacher has given us two extra credit homework assignments, and I would be eternally grateful if someone could break them down for me and show me how to do them: You are given: -Spring constant is 50N/m -Height of drop is 100m -Mass that will fall is 100kg You are asked to find the maximum length of the rope that can be used without the bungee jumper smashing into the ground. (Length of fall). ______________________________________… In the second problem, you are given: -Spring constant is 50N/m -Mass that will fall is 100kg -Length of rope is 20m You are asked to find the maximum height the bungee jumper can jump off of and not smash into the ground. Thank you!!

• Answer:

  First problem: Consider the conversion of energy. At the top, the person's kinetic energy (KE) is zero (not moving); while their gravitational potential energy (GPE) equals mgh (m=100 kg; h=100 meters). At the bottom, their KE is again zero (they have just come to a stop); their GPE is also zero (they're at the bottom); but their elastic potential energy (EPE) is ½kx² (k=50N/meter; x=amount of stretch (not yet known)); So, their GPE is converted entirely into EPE at the moment they reach the bottom. That means: mgh = ½kx² Solve for "x", plugging in the given values of "m", "h" and "k". Finally, since the rope's fully stretched length is 100 meters, we have: relaxed length + amount of stretch = 100 meters or: relaxed length = 100 meters − amount of stretch = 100 meters − x Second problem: (I think you really want to know the MINIMUM safe height, not the maximum.) Use the same equation. You have: mgh = ½kx² Furthermore, the relaxed length of the rope (L) plus the amount of stretch (x) equals the full height (h): L + x = h x = h - L Substitute: mgh = ½k(h-L)² Solve for "h", plugging in the given values for "m", "k", and "L"