Bungee jump from a bridge, how to find h

[jperez94]

Hi guys, I'm new

Now, I came here in search of help to a homework problem I have. I've tried to solve for around 40 minutes now, and can't get past solving for k. Here's the problem.

The Royal Gorge bridge over the Arkansad River is about L=310m high. A bungee jumper of mass 60kg has an elastic cord of length d=50m attached to her feet. Assume that the cord acts like a spring of force constant k. The jumper leaps, barely touches the water, and after numberous ups and downs comes to rest at a height h above the water. (a) find h. (b) find the maximum speed of the jumper

The way I'm trying to solve is is by setting up two parts. One, I just solve for K in general, so I let h equal to when the jumper touches the water, and let that be h=0. With that said, I yield U(gravity)=U(spring) (using consev. of energy, kinetic energy before the jump is =0, she has no spring potential at the top, and at the bottom she has no height, so U(gravity)=0, no kinetic as well). So that being said, I get k to equal to 121.52NM. The second part, I let my starting height equal to 260 (310 original eight minux the 50 of the bungee cord lenght), and solve for my new H, but I get an insanly big number. Am I missing something? Or am I way off in trying to solve for it? Any help is greatly appriciated. Thanks

 

[jamesrc]

For the first part, your approach is correct. You should equate:

$$mgL = \frac{k(L-d)^2}{2}$$

and solve for k. I think you should double check your arithmetic.

With the spring constant in hand, the final height h is easily found by setting the gravitational force equal to the spring force:
mg = k(L - h - d)


For the second part, the first thing we'll do is state that we're going to ignore whatever frictional effects there are to calculate the max. vel. (we know they're there, that's why it stops, but that's not important now). The maximum velocity will occur on the initial trip down at the point where the acceleration is zero (at height h. This is because the final resting place is also the equilibrium position of the system; the point about which the system oscillates). To solve the velocity as it passes this point, use the conservation of energy (again, assuming none has been dissipated yet):

$$\frac{mv^2}{2} + mgh + \frac{k(L-h-d)^2}{2} = mgL$$

 

[jperez94]

Thanks jamesrc, got it now. My problem was setting up that H, but all is well. Thanks again :)