A large cylindrical water tank 11.5 m in diameter and 13.5 m tall is supported 8.75 m above the ground by a stand. The water level in the tank is 10.6 m deep. The density of the water in the tank is 1.00 g/cm3. A very small hole is formed at the base of the vertical wall of the tank, and water is squirting out of this hole. When this water hits the ground, how far has it traveled horizontally from the hole?

Question

bobpursley · Accepted Answer

Let's apply Bernoulli's equation:

rgy1 + ½rv12 + P1 = rgy2 + ½rv22 +P2

Point 2 can be just outside the hole, so the pressure is atmospheric pressure. We're looking for v2. 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:

rgh + ½rv12 + Patm = ½rv22 + Patm

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

gh + ½v12 = ½v22

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

A1v1 = A2v2

The area of the hole is much less than the area of the cylinder, so we will simply assume that v1 is negligible compared to v2. This gives:

gh = ½v22

so v2 = [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.

yolo · Answer

13.606