A 72.0 -kg bungee jumper is standing on a tall platform ( h 0 = 42.2 m). The bun

Question

A 72.0-kg bungee jumper is standing on a tall platform (h0 = 42.2 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 = 61.8 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.

Please help! Thanks!!

Explanation / Answer

Here

Loss in potenteial Energy = Gain in Elastic potenteial Energy of the Spring

Therefore

Mg(L0 + y) = 0.5*k*y^2

72*9.8*(9.54 + y) = 0.5*61.8*y^2

y = 30.0777 m

So

required Height = h0 - (L0 + y)

= 42.2 - (9.54 + 30.0777)

= 2.5823 m

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