Posted by **Alexis ** on Thursday, November 29, 2012 at 3:48pm.

A 63.6-kg bungee jumper is standing on a tall platform (h0 = 50.6 m). The bungee cord has an unstrained length of L0 = 9.54 m and, when stretched, behaves like an ideal spring with a spring constant of k = 67.8 N/m. The jumper falls from rest, and it is assumed that the only forces acting on him are his weight and, for the latter part of the descent, the elastic force of the bungee cord. Determine how far the bungee jumper is from the water when he reaches the lowest point in his fall.

- AP Physics B -
**Damon**, Thursday, November 29, 2012 at 4:04pm
potential energy lost in fall = potential energy gained by our perfect cord

m g (fall distance) = (1/2) k (stretch distance)^2

63.6(9.81)(fall) = (1/2)(67.8)(fall-9.54)^2

623.9 f = 33.9 (f-9.54)^2

18.4 f = f^2 - 19.1 f + 91.0

f^2 - 19.1 f + 72.6 = 0

f = [ 19.1 +/- sqrt (365 - 290) ]/2

f = (19.1 +/- sqrt 75 )/2

f = (19.1 +/- 8.7) /2

= 13.9 or 5.2

5.2 is no good, shorter than cord unstretched

so

f = 13.9

50.6 - 13.9 = 36.7 above water

check my arithmetic !!!

- AP Physics B -
**Alexis **, Thursday, November 29, 2012 at 4:50pm
Thank You So Much!

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