A cylinder open to the atmosphere is full of water and standing upright on a table. There are three holes in the side of the cylinder, although these are covered to start with. One hole is 1/4 of the way down from the top, while the other two are 1/2 and 3/4 of the way down.

When the holes are uncovered, water shoots out. Which hole shoots the water furthest horizontally on the table?

- The hole closest to the top.
- The hole halfway down.
- The hole closest to the bottom.
- It's a three-way tie.

This is the subject of a homework problem, so we won't go through all the details. We will get started, however. Start with a hole a distance h from the top of the cylinder. With what speed does the water emerge from the hole?

Let's apply Bernoulli's equation:

ρgy_{1} + ½ρv_{1}^{2} + P_{1} = ρgy_{2} + ½ρv_{2}^{2} +P_{2}

Point 2 can be just outside the hole, so the pressure is atmospheric pressure. We're looking for v_{2}. Point 1 can actually be any point inside the cylinder, although some points are more convenient to work with than others. Good choices would be a point at the same level as the hole, where the pressure is atmospheric pressure + rgh, or a point at the top of the cylinder where the pressure is atmospheric pressure. Let's try a point at the top of the cylinder.

Measuring y's from the level of the hole gives:

ρgh + ½ρv_{1}^{2} + P_{atm} = ½ρv_{2}^{2} + P_{atm}

Cancelling the pressures, and then the factors of density, we're almost done:

gh + ½v_{1}^{2} = ½v_{2}^{2}

This is a good time to bring in the continuity equation:

A_{1}v_{1} = A_{2}v_{2}

The area of the hole is much less than the area of the cylinder, so we will simply assume that v_{1} is negligible compared to v_{2}. This gives:

gh = ½v_{2}^{2}

so v_{2} = [2gh]^{½}

This should look familiar to you.

The rest of the analysis involves recognizing that the water emerges from the hole with an initial horizontal velocity, and applying projectile motion equations to determine the horizontal distance reached by the water stream.