A 63.0-kg bungee jumper is standing on a tall platform (h0 = 46.0 m), as indicat

Question

A 63.0-kg bungee jumper is standing on a tall platform (h0 = 46.0 m), as indicated in the figure. The bungee cord has an unstrained length of L0 = 9.00 m, and when stretched, behaves like an ideal spring with a spring constant of 64.0 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.

Picture: http://i14.photobucket.com/albums/a322/AznHurricanesUM/EAT_12277849565910_6382961148501805.jpg

Explanation / Answer

m = 63.0-kg, h0 = 46.0 m, L0 = 9.00 m, k = 64.0 N/m. Find the distance x between the bungee jumper and the water when he reaches the lowest point in his fall. The final length of the cord = h0 - x
the cord stretches h0 - x - L0 elastic potential energy = k(h0 - x - L0)^2/2 initial gravitational potential energy = mgh0 final gravitational potential energy = mgx
energy conservation: mgh0 = mgx + k(h0 - x - L0)^2/2 2mgh0 = 2mgx + k(h0 - x - L0)^2 (h0 - x - L0)^2 + 2mgx/k - 2mgh0/k = 0
(37 - x)^2 + 19.29x - 887.5 = 0
let 37 - x = y y^2 + 19.29(37 - y) - 887.5 = 0 y^2 - 19.29y - 173.8 = 0
y = 26 m
so x = 37 - 26 = 11 m

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