Bungee Jumping Physics Question?

C

chickenhouse

New member
What is the formula for determining how close to the ground a bungy jumper will be at maximum stretch when you have the following variables?

Height of bungee tower: H
height of jumper: h
Spring constant of cord: k
Weight of jumper: w
Length of cord: L

Also if you could tell me a likely actual value for k it would be appreciated.

Thanks
 
Use conservation of energy (which you can use, because only gravity and the cord perform work): the energy in the top of the tower is equal to the energy in the maximum stretch point.

I'd say a likely actual value for a bungy cord would be a few hundreds of N/m.
 
I guess it would depend on how you want to look at it. I'm gonna say that the top of the relaxed spring is where height = 0. The person's initial height is at h at phase one. Then at phase 2 she touches the spring (I drew the spring below the bunjee) at which point there is only kinetic energy since there is no deformation in the spring. Then at the final phase the spring is compressed at a distance of " s"

I <--- This is "h"
I
I
I
....... <--- This is h = 0 the reference point
I
I <--- This is the compresed length s
-------
I
I <--- This is the length below the compressed distance 's' call it l

1/2 x mv^2 = mg(-s) + 1/2 x ks^2
s = (mg +/- (mg^2 + 2mghk)^(1/2)/k)

Total distance is s + l and w = mg

s = (w +/- (w^2 + 2whk)^(1/2)/k)

Sorry this is the best my brain can do ... LOL
 
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