Bungee Jump (link for coaches): https://phy-coaches.cst.cmich.edu/type3/E8-bunge
ID: 1462363 • Letter: B
Question
Bungee Jump (link for coaches): https://phy-coaches.cst.cmich.edu/type3/E8-bungeejump/bungeejump.php You have a summer job working for a company that arranges bungee jumps in which the jumpers step off a high platform with a bungee cord attached to their legs. The cord used in the jump are sorted by their unstreched length and elasticity. Your first task is to help the employees pick out cords by developing an equation for the unstreched length of a cord that is appropriate for a particular jump, given the mass of the jumper m 68 kg and the height of the jump h-110 m. For most exciting jump, the person should stop just short of the ground. In order to keep the jumper safe, the company doctor recommends that the maximum acceleration of the person during the jump be no longer than 2g. Theofa Bungee Jump (link for coaches): https:/lphy-coaches.cst.cmich.edu/type3/E8-bungeejump/bungeejump.php anges Youper commExplanation / Answer
You can see the bungee rope as a spring which obey the Hooke's Law |F| = k x, where x is how long stretch the spring (or the rope) and is a constant related to the spring (rope) easticity.
When someone is hanging with the rope, the rope will stretch until the force made by itself get equal to the weigth of the person
kx = mg --> x = mg/k, it will the new equilibriun possition for the system person - rope.
Now, if you move the person out of the equilibriun point and release it, it will oscilate arround that equilibriun point with an amplitude equal to the displacement about the equilibriun point. So the max length of the rope L for a jump of heigth H must be:
H = 2 (L + mg/k) --> L = (H/2) - (mg/k)
Now, the maximal acceleration will occur when the rope excert the maximal force, it will occur when the rope gets its maximal stretch Fmax = k [(H/2) + (mg/k)] = (kH/2) + mg
So that, amax = Fmax/m --> amax = [(kH/2) + mg]/m = (kH/2m) + g --> k = (amax - g) (2m/H).
Finally:
L = (H/2) - (mgH/[(amax - g)2m])
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