Velocity of bungy jumper, very confusing?

[x86]

Homework Statement

A bungee jumper with a mass of 78kg tied to a 39 m cord jumps off a bridge from a height of 69 m. She falls to 6 .0m above the ground before the chord brings her momentarily to rest. Calculate the impulse exerted on the jumper by the cord as it stretches

Homework Equations

Ft = mvf - mvi
P = mv
Eg = mgh
Ep = 1/2 x k^2
Ek = 1/2 m v ^2

The Attempt at a Solution

Impulse = 0 - mvi

To find vi:

Ek1 + Eg1 = Eg2 + Ep2

So here is where I'm stuck. The equation is impossible to solve because there are two unknowns (spring constant and velocity)

 

[rock.freak667]

Why would you need the spring constant for? Once you have the velocity, you can get the change in momentum as you have the mass.

 

[x86]

Why would you need the spring constant for? Once you have the velocity, you can get the change in momentum as you have the mass.

I need the spring constant because some energy is converted into elastic potential energy.

Ek1 + Eg1 = Eg2 + Ep2

That is the equation I've made. LS = the instant before falling, RS = on stop

So it comes out to 0.5mv^2 = mgh = mgh + 0.5kx^2

In order to find the velocity I have to have k but its an unknown. Am I doing this right? Did I set it up correctly? I know there has to be some potential energy because the rope is only 39 m long and it stretches far beyond that point as the person falls around 69 m - 6m and the rope isn't that long

 

[TSny]

Actually you won't need the spring constant to find the impulse of the cord. You wrote

impulse = 0 - mv_i.

This will be the key! Think about the meaning of v_i. What point of the jump does this refer to?

Also, for the left side of the equation do you think that represents the impulse of the cord, the impulse of the force of gravity, or the total impulse of all of the forces acting on the jumper?

 

[TSny]

Oops, You might indeed need the spring constant. But you should be able to find it.

 

[TSny]

I need the spring constant because some energy is converted into elastic potential energy.

Ek1 + Eg1 = Eg2 + Ep2

That is the equation I've made. LS = the instant before falling, RS = on stop

What is the velocity at "the instant before falling"?