Max speed and tension of bungee jump?

[Arin]

Just signed up, hi everyone!

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:
T₁+V₁=T₂+V₂
Elastic potential energy = 0.5*k*x², in this case = 250*x^{^2}

The Attempt at a Solution

To find length of unstretched cord,
Conservation of energy:
T₁+V₁=T₂+V₂
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.

 

[Arin]

Bump!

 

[gneill]

Hi Arin, Welcome to Physics Forums.

When you have the position for the maximum velocity (L + x) you should be able to find the kinetic energy for that instant. All the energy originates from the change in gravitational PE to that point, some of it being transformed into elastic PE on the way. What's left must be the KE.

 

[Arin]

I figured it out, thread closed!