Bungee Jumper - Elastic Potential Energy

[aeromat]

Homework Statement

In a "head dip" bungee jump from a bridge over a river the bungee cord is fastened to the jumper's ankles. The jumper than steps off and falls towards the river until the cord becomes taut. At that point, the cord begins to slow the jumper's decent, until his head just touches the water. The bridge is 22.0m above the river. The unstretched length of the cord is 12.2m. The jumper is 1.80m tall and has a mass of 60.0kg. Determine the

a) Required value of the string constant for this jump to be successful.

b) Acceleration of the jmper at the bottom of the decent

Homework Equations

Ee = 1/2(k)(x)^2

The Attempt at a Solution

I know x = 12.2m, since that is the the length of the string.
I am using the gravitational potential energy of the start point being at the top of the bridge, and here the total height is 1.80 + 22.0m which is 23.8m; taking into account his height and the bridge together.

Thus, I calculated Eg = (60.0)(9.81)(1.80 + 22.0)
Eg = 14008.68 J

At the final point, Eg' = 0, Ee is not equal to 0.

Eg = Ee'
(14,008.68J) = 1/2(k)(12.2)^2
$$k = 1.88*10^2 N/m$$

However, the back of the textbook has the following k value: 405N/m. What am I doing wrong?

 

[berkeman]

Homework Statement

In a "head dip" bungee jump from a bridge over a river the bungee cord is fastened to the jumper's ankles. The jumper than steps off and falls towards the river until the cord becomes taut. At that point, the cord begins to slow the jumper's decent, until his head just touches the water. The bridge is 22.0m above the river. The unstretched length of the cord is 12.2m. The jumper is 1.80m tall and has a mass of 60.0kg. Determine the

a) Required value of the string constant for this jump to be successful.

b) Acceleration of the jmper at the bottom of the decent

Homework Equations

Ee = 1/2(k)(x)^2

The Attempt at a Solution

I know x = 12.2m, since that is the the length of the string.
I am using the gravitational potential energy of the start point being at the top of the bridge, and here the total height is 1.80 + 22.0m which is 23.8m; taking into account his height and the bridge together.

Thus, I calculated Eg = (60.0)(9.81)(1.80 + 22.0)
Eg = 14008.68 J

At the final point, Eg' = 0, Ee is not equal to 0.

Eg = Ee'
(14,008.68J) = 1/2(k)(12.2)^2
$$k = 1.88*10^2 N/m$$

However, the back of the textbook has the following k value: 405N/m. What am I doing wrong?

I didn't go through your numbers much yet, but the initial height of the center of mass of the idiot, er, I mean jumper, is not 22m + 1.8m. What should it be? And where is that center of mass when his head clips the water?

 

[aeromat]

Ok so I know that the height is just the bridge itself, so I set it to this scenario:

Eg[before jump] = Ee [upon reaching lowest point, where the Eg would equal 0]
(m)(g)(h) = 1/2(k)(x)^2