Efflux Jets from a Tank

A tank is kept full to 1.20 m. Three holes are drilled in the side of it, 0.30 m, 0.60 m and 0.90 m above the table it stands on. Work out where each jet lands, and find the height that throws the water furthest.

1. Water leaving a hole has been pushed by the depth above it and by nothing else. That pressure buys it exactly the speed a dropped stone would have after falling the same distance, which is Torricelli's result. The lowest hole has 0.90 m of water over it.

   $$v=\sqrt{2gh}=\sqrt{2\times 9.81\times 0.90}=4.20m/s$$

2. Out in the air the jet is a projectile fired horizontally, so how long it stays up is decided by how far it has to fall. The depth above the hole sets the speed. The height below the hole sets the time.

   $$t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2\times 0.30}{9.81}}=0.2473\mathrm{s}$$

3. Multiply the two and g cancels out of the answer. A jet on the Moon leaves slower and hangs in the air longer, and it lands on the same spot.

   $$R=vt=\sqrt{2gh}\sqrt{\frac{2y}{g}}=2\sqrt{hy}=2\sqrt{0.90\times 0.30}=1.039\mathrm{m}$$

4. The range only ever sees the product of the depth above and the height below, and those two always add up to H. Swapping y for H − y leaves the product alone, so the hole at 0.90 m throws exactly as far as the hole at 0.30 m.

   $$R(y)=2\sqrt{(H-y)y}⟹R(0.90)=2\sqrt{0.30\times 0.90}=1.039\mathrm{m}=R(0.30)$$

5. Two numbers with a fixed sum multiply to the most when they are equal, which puts the best hole halfway up. Depth above and height below are then both H/2, and the range comes out at H on the nose.

   $$\underset{y}{\max }(H-y)y⟹y=\frac{H}{2}=0.60\mathrm{m},R_{\max }=2\sqrt{\frac{H}{2}\cdot \frac{H}{2}}=H=1.200\mathrm{m}$$

6. So no tank throws water further along the table than it is deep, whatever you do with the drill. And the twins land together without arriving together: the upper jet is airborne for 0.4284 s against the lower one's 0.2473 s, so the wet spot is hit twice, 0.18 s apart.

   $$t_{0.90}=\sqrt{\frac{2\times 0.90}{9.81}}=0.4284\mathrm{s},t_{0.90}-t_{0.30}=0.1810\mathrm{s}$$