The stream of water is fatter near the mouth of the faucet, and skinnier lower down. This can be understood using conservation of mass. Since water is being neither created nor destroyed, the mass of the water that leaves the faucet in one second must be the same as the amount that flows past a lower point in the same time interval. The water speeds up as it falls, so the two quantities of water can only be equal if the stream is narrower at the bottom.

Assume the velocity of water is v(0),
the speed will increase as it fall a height h, and v(h)=v(0)+\sqrt(2 g h), where g is the gravity.

The amount of water is proportional to v(h)A(h)=v(h) \pi r^2(h) which is a constant.
When the water speed , it's radius become smaller.
Assume r(0) is the initial radius, then r(h)=\frac{r(0)}{\sqrt{v(0)+2 g h}}

Modeling activity:
Try to drag the yellow square to fit with the shape of the water stream, by adjusting initial radius r(0)of the stream and initial speed v(0) of the water.

translate strings in simulation to different language format before download
Full screen applet or Problem viewing java?Add to exception site list
Press the Alt key and the left mouse button to drag the applet off the browser and onto the desktop. This work is licensed under a Creative Commons Attribution 2.5 Taiwan License
Download EJS jar file(982.4kB):double click downloaded file to run it. (13 times by 13 users) , Download EJS source View EJS source