Bungee Jumping at 1,000 Feet: What Happens?

[john gault]

You are in a hot air balloon at 1,000 feet altitude with you, a pilot and a bungee cord with one end attached to your leg and the other end attached to the basket of the balloon. You have found a point of absolute neutral bouyancy and there is no wind or any other change in the environment from now on. You then jump out of the balloon with the bungee cord attached and fall as in any other bungee jump (until the cord stretches taut, you "bottom out", bounce back, fall again, etc.).

Once you are slack, hanging still by the ankle 50 below the basket, is the balloon higher, lower or at the same 1,000 foot altitude you were before jumping? Assume the bungee cord weighs 20 lbs and you weigh 180 lbs.

-JG

 

[whozum]

This is a poorly written problem. I'll give you a hint but I need your input, the weight of everything (balloon, cord, you) is constant, nothing ispermanently removed from the system, so you know the total weight is the same. Although the man is no longer sitting in the balloon, he is still pulling it down with hisweight by hanging from it.

Somthing interesting happens while he's falling though, what is it?

 

[jdavel]

whozum,

But the air density is slightly higher where the jumper ends up than it was where he started.

 

[pallidin]

Actually, the balloon first moves up, then is dragged down, and the elasticity of the bungee rope with a weight(you) causes this dampening cycle to continue until an equilibrium is reached.

 

[arun_mid]

If it's possible to assume that the man-balloon system is in equilibrium throughout the process(this would be true if the system was initially in equilibrium--is this what is meant by 'point of absolute neutral buoyancy'? Also, the density change of air should be neglected), the CM position is unchanged. I think that a 50ft ht difference is quite small, so the assumption is ok. Then the problem can be solved by taking the final distance of the man from his initial level as x(taking the initial level as the origin), and finding the final position of the center of mass in terms of x using the definition of CM(you would have to use a bit of calculus on the bungee cord, but I am not familiar with typing stuff like integral signs, so I can't provide the working). Then equate this to zero, as the CM is at the initial level, which is taken as the origin.

 

[jdavel]

But if you ignore the variation of density with height, then the balloon/man system has neutral buoyancy everywhere. The equilibrium is unstable, and the problem really becomes indeterminant.

 

[arun_mid]

Sorry for the late reply. I don't understand what jdavel means by the equilibrium being unstable, especially after saying that the system has neutral buoyancy everywhere--which is true if you take the density constant everywhere. This is not the case, but it can be taken as approximately constant over the maximum distance through which the man bungee jumps (probably larger, but not much larger, than 50ft).

 

[russ_watters]

Sorry for the late reply. I don't understand what jdavel means by the equilibrium being unstable, especially after saying that the system has neutral buoyancy everywhere--which is true if you take the density constant everywhere.

If the density is the same everywhere, then a balloon that is rising will rise forever and a balloon that is falling will fall forever. That's why its an unstable situation.

However since the density of the balloon is relatively low (on the same order of magnitude as the atmosphere) and the density of the man is relatively high (several orders of magnitude above that of the atmosphere), you can ignore the buoyancy/altitude of the man and only consider the buoyancy/altitude of the balloon unless the OP is really looking for an answer in inches, which I doubt.

So the answer is the balloon is at the same altitude - unless you want to be really specific, in which case its a couple of inches higher.

 

[whozum]

If the density is the same everywhere, then a balloon that is rising will rise forever and a balloon that is falling will fall forever. That's why its an unstable situation.

However since the density of the balloon is relatively low (on the same order of magnitude as the atmosphere) and the density of the man is relatively high (several orders of magnitude above that of the atmosphere), you can ignore the buoyancy/altitude of the man and only consider the buoyancy/altitude of the balloon unless the OP is really looking for an answer in inches, which I doubt.

So the answer is the balloon is at the same altitude - unless you want to be really specific, in which case its a couple of inches higher.

Wouldn't it oscillate about the equilibrium level for a while?

 

[arun_mid]

I didn't mean that the density of air was constant everywhere. But assuming that the variation with height is low, I don't think that the difference in density between the height of the balloon and that of the man at any instant would be considerable. Anyway, if it IS it leads to a lot of complications. (Anyhow, isn't the case of the balloon moving in air of uniform density one of NEUTRAL equilibrium? Because, in this case the balloon doesn't accelerate at any instant of its motion.)
Has anyone come up with a method that takes into account variation of density over that height difference? Moreover, if you take this into account, you would probably be unable to come up with a formula for the variation because air is not even static. Any approach would have to ignore or be INDEPENDENT of this density variation. Any ideas? But please tell me if there are any other drawbacks to my method.
As for the oscillations about the equilibrium level, the question doesn't clearly state whether the balloon comes to the equilibrium level with zero velocity, if not, the man/balloon system will oscillate. that makes the problem quite complicated indeed.