Conservation of Energy: Bungie Jumping

[Seda]

Homework Statement

Bungie Jumper starts 65 meters from ground.
Wants to end fall 10m from ground (55 meters from the top).
Cord follows Hookes law.

We know that when his body weight hangs at rest from a 5 meter section of cord, he stretches the cord 1.5 meters.

How long should his cord be?

Homework Equations

Gravitational Potential Energy = mgh
Elastic potential energy = .5kx^2

The Attempt at a Solution

Well, we know that mg=k(1.5) in reference to the last part of the question, because he would be hanging from the cord at rest [no net force]. We'll use this later.

All of his initial energy before the actual jump is gravitational potential energy so ENERGY = mg*65.

When, he is 10m from the ground (at the end of the jump), he has some Gravitational and some elastic potential energy, so ENERGY = mg*10 + .5kx^2.

Setting the two ENERGIES equal, I get a final equation of:
mg*65 = mg*10 + .5kx^2.

and using m=1.5k/g (from first part) I get:
(1.5k)*65 = (1.5k)*10 +.5kx^2

all k's cancel, and I can solve for x.

I get x=12.85. Subtracting this from the total jump distance of 55m, gives me a cord length of 42.15m.

The correct answer is supposedly 25.8m.

Can someone point out what I am doing wrong?

 

[IBY]

Are you sure that is the correct answer? I got the same answer as you did.

 

[Seda]

Are you sure that is the correct answer? I got the same answer as you did.

Well, it's the answer given in the book, it may be a typo. I wanted to make sure I wasn't doing anything stupid. Thanks.