Pierce a tall plastic bottle a quarter, a half and three quarters of the way down. Fill it with water and stand it on a table. Three jets spring out. The bottom jet is the fastest, since it has the most water above it. Yet it is not the one that lands farthest. The middle jet reaches farthest, and the top and bottom jets strike the table together.
Torricelli meets the projectile
A tank stands on level ground and holds water to a depth \(H\). A small hole in its side is at a depth \(h\) below the surface, so it is \(H - h\) above the ground. The water leaves horizontally at \(\sqrt{2gh}\). From that instant each bit of water is a projectile thrown horizontally, as in the chapter on Motion in Two Dimensions. It keeps its horizontal speed and falls freely through the height \(H - h\).
\[ v = \sqrt{2gh} \]
Why. Torricelli's law gives the horizontal speed of the jet.
\[ t = \sqrt{\dfrac{2(H - h)}{g}} \]
Why. the water falls from rest vertically through \(H - h\), so \(H - h = \tfrac12 g t^2\).
\[ R = vt = \sqrt{2gh}\,\sqrt{\dfrac{2(H - h)}{g}} \]
Why. the horizontal speed is constant, so the range is the speed times the time of fall.
\[ R = 2\sqrt{h(H - h)} \]
Read it. \(g\) cancels under the square root, leaving \(\sqrt{4h(H - h)}\).
The range does not contain \(g\). A deeper hole gives a faster jet, and a stronger gravity gives a faster jet too. But stronger gravity also brings the water down sooner, and the two effects cancel exactly. Neither the density of the liquid nor the size of a small hole enters the speed or the fall, so neither changes the range.
A tank on the ground holds water to a depth of \(5\,\text{m}\). A small hole in its side is \(1.8\,\text{m}\) below the surface. Take \(g = 10\,\text{m/s}^2\). Find the speed of the jet, its time of fall and the range.
\(H = 5\,\text{m}\) and \(h = 1.8\,\text{m}\), so the hole is \(H - h = 3.2\,\text{m}\) above the ground.
\(v = \sqrt{2gh}\), \(t = \sqrt{2(H - h)/g}\), \(R = vt\).
\(v = \sqrt{36} = 6\,\text{m/s}\), \(t = \sqrt{0.64} = 0.8\,\text{s}\), \(R = 6 \times 0.8 = 4.8\,\text{m}\).
The water leaves at \(6\,\text{m/s}\), is in the air for \(0.8\,\text{s}\), and lands \(4.8\,\text{m}\) from the tank.
A tank on the ground holds water \(2.5\,\text{m}\) deep. A small hole is \(0.9\,\text{m}\) below the surface. How far from the tank does the jet land, in m?
Two holes, one landing point
Put \(H - h\) in place of \(h\) in the range. The product \(h(H - h)\) becomes \((H - h)h\), the same. So a hole at depth \(h\) and a twin hole at depth \(H - h\) throw their water to the same spot. In the sim above, the dashed jet always meets the solid one on the ground.
The two jets reach the same point in different ways. The upper hole has little water above it, so its jet is slow, but it is high up and falls for a long time. The lower hole has a fast jet with a short fall. In the \(5\,\text{m}\) tank, holes \(1\,\text{m}\) and \(4\,\text{m}\) down both land \(2\sqrt{1 \times 4} = 4\,\text{m}\) away. The pair at \(1.8\,\text{m}\) and \(3.2\,\text{m}\) both land at \(4.8\,\text{m}\). The upper jet takes \(0.8\,\text{s}\), and the lower one only \(0.6\,\text{s}\).
Both twins also strike the ground with the same speed. Every bit of water started at rest at the surface and ends \(H\) lower, with no energy lost. So its landing speed is \(\sqrt{2gH}\) for any hole: \(10\,\text{m/s}\) in the \(5\,\text{m}\) tank. For the hole at \(1.8\,\text{m}\), the horizontal part is \(6\,\text{m/s}\) and the vertical part \(gt = 8\,\text{m/s}\), and \(\sqrt{36 + 64} = 10\,\text{m/s}\).
The hole that throws farthest
Which depth gives the greatest range? \(R^2 = 4(Hh - h^2)\) is largest when its slope in \(h\) is zero: \(H - 2h = 0\), so \(h = H/2\). The greatest range is then \(2\sqrt{(H/2)(H/2)} = H\). A hole halfway down throws the water as far as the water is deep. In the sim, the graph of \(R\) against \(h\) peaks at \(h = 2.5\,\text{m}\) with \(R = 5\,\text{m}\), and it is symmetric about that depth.
To find the depth for a given range, square the range formula. \(R^2 = 4h(H - h)\) is a quadratic, \(h^2 - Hh + R^2/4 = 0\), with two roots that add up to \(H\). They are the twin holes. For the \(5\,\text{m}\) tank and \(R = 4\,\text{m}\), \(h^2 - 5h + 4 = 0\) gives \(h = 1\,\text{m}\) or \(4\,\text{m}\). A range greater than \(H\) gives no real root: no hole can reach it.
A tank on the ground holds water \(6\,\text{m}\) deep. A hole \(1.5\,\text{m}\) below the surface throws its jet to a certain point. At what other depth could a hole throw its jet to the same point?
A tank on the ground holds water \(3.6\,\text{m}\) deep. What is the greatest range a jet from a small hole in its side can reach, in m?
Two holes in a tank on the ground are at depths \(h\) and \(H - h\), and their jets land at the same point. How do their speeds just before they hit the ground compare?
The path of the jet
Take the hole as the origin, with \(x\) horizontal and \(y\) measured downward. After a time \(t\) the water is at \(x = vt\) and \(y = \tfrac12 g t^2\). Eliminate \(t\) with \(t = x/v\) and put \(v^2 = 2gh\).
\[ y = \dfrac{g x^2}{2v^2} = \dfrac{g x^2}{2 \times 2gh} = \dfrac{x^2}{4h} \]
The jet is a parabola whose shape depends only on the depth of the hole. Suppose a hole is \(0.8\,\text{m}\) deep and a wall stands \(1.6\,\text{m}\) from the tank. The jet has dropped \(1.6^2/(4 \times 0.8) = 0.8\,\text{m}\) below the hole when it strikes the wall.
A hole is \(1.25\,\text{m}\) below the surface of the water in a tall tank. A wall stands \(2\,\text{m}\) from the side of the tank. How far below the level of the hole does the jet strike the wall, in m?
A platform, a tilted nozzle, a lift
The range formula assumed that the jet lands at the level of the tank's base. If the tank stands on a platform of height \(d\), the jet falls \(H - h + d\) instead. The same steps give \(R = 2\sqrt{h(H - h + d)}\). Then the hole halfway down no longer throws farthest, because the extra fall favours faster jets from deeper holes.
On the platform in Fig 5.1 the hole is \(3.2 + 1.8 = 5\,\text{m}\) above the ground. The jet falls for \(\sqrt{2 \times 5/10} = 1\,\text{s}\) and lands \(6 \times 1 = 6\,\text{m}\) away. The formula agrees: \(2\sqrt{1.8 \times (5 - 1.8 + 1.8)} = 2\sqrt{9} = 6\,\text{m}\).
A nozzle can also point the jet upward at an angle \(\theta\). The water still leaves at \(\sqrt{2gh}\), but now it is a projectile thrown at an angle. Split the speed into \(\sqrt{2gh}\cos\theta\) across and \(\sqrt{2gh}\sin\theta\) up, and treat the two motions separately. In the third chip of Fig 5.1 the jet leaves at \(6\,\text{m/s}\) at \(37°\), with parts \(4.8\,\text{m/s}\) across and \(3.6\,\text{m/s}\) up. It rises \(3.6^2/20 \approx 0.65\,\text{m}\) above the nozzle before it falls to the ground.
A garden fountain is fed from a tank whose water surface is \(5\,\text{m}\) above the nozzle. The nozzle is at ground level and points \(37°\) above the horizontal. Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\) and \(\cos 37° = 0.8\). Find how high the water rises and how far from the nozzle it lands.
The water leaves at \(\sqrt{2gh} = \sqrt{100} = 10\,\text{m/s}\), with parts \(8\,\text{m/s}\) across and \(6\,\text{m/s}\) up.
It rises \(6^2/(2 \times 10) = 1.8\,\text{m}\) and is in the air for \(2 \times 6/10 = 1.2\,\text{s}\).
It lands \(8 \times 1.2 = 9.6\,\text{m}\) away, as the range formula \(u^2\sin 2\theta/g = 100 \times 0.96/10\) confirms.
Now put the tank in a lift that accelerates upward at \(a\). In the chapter on Pressure in Fluids, the water in such a lift behaves as if gravity were \(g + a\). The jet then leaves faster, \(\sqrt{2(g + a)h}\). But seen from the lift, it also falls faster, with \(t = \sqrt{2(H - h)/(g + a)}\). The product is again \(2\sqrt{h(H - h)}\), and the range does not change. The same holds on the Moon, where both are slower.
A tank of water with a hole in its side stands on the floor of a lift. The lift starts to accelerate upward. Seen from inside the lift, what happens to the point where the jet lands on the floor?
Use \(R = 2\sqrt{h(H - h)}\) only when the jet lands at the level of the base of the tank. On a platform, or with the jet landing below the base, find the time from the actual fall. Then multiply by \(\sqrt{2gh}\).
Now return to the bottle with three holes, standing on a table with depth \(H\). The holes are at \(H/4\), \(H/2\) and \(3H/4\) below the surface. The middle hole is the best depth, \(H/2\), so its jet lands farthest, a distance \(H\) away. The top and bottom holes are twins, since \(H/4 + 3H/4 = H\). Both land at \(2\sqrt{(H/4)(3H/4)} \approx 0.87H\), at the same spot.
- Using the height of the hole above the ground in \(v = \sqrt{2gh}\).The speed comes from the depth below the surface; the height above the ground gives the time of fall.
- Thinking the lowest hole always throws farthest.A deeper hole gives a faster jet but a shorter fall. The best hole is halfway down.
- Using \(R = 2\sqrt{h(H - h)}\) for a tank on a platform.Add the platform height to the fall: \(R = 2\sqrt{h(H - h + d)}\).
- Expecting the range to change with \(g\), in a lift or on the Moon.\(g\) cancels. The jet is faster in stronger gravity, but it also falls sooner.