# If a 100 kg swimmer drops in a pool, how far down will he go?

Imagine that a trap door quickly opens and a 100 kg (2 meter tall) man IMMEDIATELY drops into the water (STP)-- the waterline is at is his feet, and so there is no distance between his feet and the water before the trap door opens. His hands and feet are not propelling him in anyway BTW, and the water is not brine. How deep will his feet reach into the bottomless (for this experiment) depths?

Assume that the man had held a full breath of air ( in an average set of lungs ) as he took the plunge . How far down will he descend ?

Of course he drops vertically, with legs together. The problem must be simplified where it can don't you agree?

### 1 Answer

- simplicitusLv 71 decade agoFavorite Answer
Does he go vertically, does he belly flop, does he scrunch up as in a cannonball, or does he point his toes, keep his legs together, etc. so as to minimize drag? Or are we assuming he is 2 meter cylinder of the same overall density as a 100kg man? And if the latter, what is his BMI? (Fatty folk have lower density than lean ones)

What assumptions are we making about water friction? Is this just an gravity - buoyancy question?

If we assume completely unrealistically, a cylinder of uniform density, and no friction, etc. then the depth can be computed using energy considerations.

His initial energy is entirely potential energy due to gravity - equal to mgh, with h = the height of his center of gravity above the water. His final energy is also entirely potential - due to the buoyancy of the water.

The easiest way to think of it is that he now has negative mass. That is mass submerged, Ms = mass in air - mass of displaced water.

And the distance down from the surface, Hs, is negative height. So we have:

Ms Hs = m h

We are given m and h, we can get a guess of the volume and determine Ms, so we can solve for Hs.

Of course, one can't really ignore friction, so this over-estimates the depth significantly. (In fact, without friction, the buoyancy from his depth will propel him back to his starting position.)

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