- #1

Arin

- 4

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**Just signed up, hi everyone!**

1. Homework Statement

1. Homework Statement

A man weighs 150 lb, and attaches a bungee cord having a stiffness of k = 500 lb/ft, to his feet.

If he jumps from rest off the side of a bridge, determine the required unstretched length of the

cord so that he can just touch the surface of the water 120 ft below when he reaches the end of

his fall. Also, compute the maximum tension in the cord and his maximum speed. Neglect his

size in the calculation.

Given:

Weight=mg=150 lbs

k=500 lb/ft

Bridge height = 120 ft

Find:

Length of cord

Max tension of cord

Max speed

## Homework Equations

Conservation of energy:

T

_{1}+V

_{1}=T

_{2}+V

_{2}

Elastic potential energy = 0.5*k*x

^{2}, in this case = 250*x

^{^2}

## The Attempt at a Solution

To find length of unstretched cord,

Conservation of energy:

T

_{1}+V

_{1}=T

_{2}+V

_{2}

0 + (150*120) = 0 + 250*x^2

x=sqrt(72)

Length of cord (L)

120=L+sqrt(72)

L=111.515 ft

From reading this thread - https://www.physicsforums.com/threads/the-physics-behind-bungee-jumping.16539/ - Looking at Chen's post on first page...

I have attempted to solve for max speed by thinking that max speed is attained at the point where he has a minimum in total potential energy

Σpotential energy = E

_{grav}+E

_{elastic}= mgh+250*x^2

This is the part where I'm lost,

I tried setting up:

Σpotential energy = 150 (120-sqrt(72)-x)+250 x^2

X=0.3 when minimal

Not sure if this is right at all... Was going to use:

ΣPE = mg(200 - x) + (1/2)*k*x^2

But this doesn't seem like it's right because it's using the first x as total distance from bridge and also x as how far the bungee stretches...

After finding the right x (stretch length when max velocity), I would use conservation of energy to find a speed at this point?

And then tension would be simply 150 right, because at the point he's still at the bottom, the rope would be holding him?

Please help! Thanks. Been working on this for a while.