# BUNGEE JUMPING,Problem is SEEMS long but it is not

Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord will first straighten and then stretch as Kate falls. Assume that the bungee cord behaves as an ideal spring once it begins to stretch. Neglect the effects of air resistance and use g for the magnitude of the acceleration due to gravity. Kate wants to adjust the length of the bungee cord so that she can come as close as possible to the river when she jumps, without actually hitting the water. Kate decides that she needs to measure the effective spring constant of the bungee cord,k . She attaches herself to the cord and asks a couple of strong friends to lower her slowly from the bridge until she can just touch the water (at this point the cord is stretched). She signals to her friends who mark the cord where they are holding it and haul her back up to the surface of the bridge. They measure the length of the (unstretched) bungee cord from the point where it was marked to the end attached to Kate. The length they measure is L1.

and we know that k=m*g/(h-L1)

QUESTION)Now that Kate knows the spring constant of the bungee cord, what (unstretched) length of cord,L2 , should she give herself when she jumps off the bridge? (Remember that she wants to get as close as possible to the water without getting wet.)

For simplicity, assume that Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward. Also, assume that her height is negligible compared to the length of the bungee cord and that the bungee cord has the same spring constant in parts A and B.

Express L2 in terms of L1 and any quantities given in the problem introduction.

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Let the amount of stretch in the chord be x, and if L2 is the unstretched length of the string, we need ...

#### Solution Summary

A bungee jumping problem is analyzed. The quantities for problem introduction is provided.