A Large Open Tank Has Two Holes

Picture this: you're standing next to a big, open tank filled with water. Someone pokes two holes in the side—one near the top, another closer to the bottom. Then they ask you: which stream shoots farther?

Most people guess the top hole. Practically speaking, it seems like the water should squirt out with more force from higher up. But here's the thing—nature doesn't always play by our intuition. The real answer has to do with something called exit velocity*, and it turns out the hole lower down wins every time.

What Is This Problem, Really?

This isn't just a brain teaser—it's a classic fluid dynamics problem that shows up in physics classes, engineering exams, and even job interviews at companies that work with liquids. At its core, it's asking: given two holes at different heights in a vertical tank, which one sprays water farther horizontally?

The setup is deceptively simple. In practice, you've got a large open tank—meaning the water level stays roughly constant as it drains—and two small holes punched through the side. Gravity pulls the water down, pressure pushes it out through the holes, and then the stream follows a parabolic path to the ground.

What makes this interesting is that the answer depends on two competing factors: how fast the water shoots out, and how long it has to fall before hitting the ground.

Why This Matters

Understanding this problem helps with way more than just academic curiosity. It's the foundation for calculating things like:

How far a bathtub drain will pull objects
The range of a garden sprinkler
How fluids behave in plumbing systems
Even the physics behind some types of water features

But more importantly, it teaches you to think about physics problems in terms of trade-offs. It's not just about speed or just about height—it's about how they work together.

The Physics Behind the Spray

Let's break this down step by step, because the math here is actually pretty elegant once you see how it pieces together.

Pressure and Exit Velocity

The water doesn't just magically shoot out of the holes. And there's a pressure difference driving it. At any point inside the liquid, the pressure comes from two sources: atmospheric pressure pushing down on the surface, and the weight of the water above that point.

The deeper you go, the more water is pressing down, so the pressure increases linearly with depth. This means the hole closer to the bottom experiences higher pressure—and therefore higher velocity—as the water exits.

We can calculate that exit velocity using Torricelli's law, which says the speed of efflux (that's a fancy word for "water coming out") equals the square root of twice the height of the water column above the hole times gravity. In formula terms: v = √(2gh), where h is the depth below the surface.

So if one hole is twice as deep as the other, it doesn't shoot twice as fast—it shoots about 41% faster, because of that square root relationship.

Time of Flight

Here's where it gets interesting. On top of that, the water doesn't just zoom horizontally forever. It also falls downward under gravity. The lower hole has less distance to drop before reaching the ground, which means less time in the air.

The time it takes to fall depends on how high the hole is above the ground. If we call that height H, then the time to fall is t = √(2H/g), again using basic kinematics.

A lower hole means a smaller H, which means a smaller t. The water spends less time flying through the air.

Horizontal Distance

Now we combine these two effects. Horizontal distance equals horizontal velocity times time of flight. So:

d = v × t = √(2gh) × √(2Hg)

When you work through the algebra, this simplifies to d = 2√(hH).

Wait, let me make sure I got that right. Actually, let me recalculate:

d = √(2gh) × √(2Hg/g) = √(2gh) × √(2H/g) × √g = √(2gh) × √(2H) = √(4ghH) = 2√(ghH)

Hmm, that's not quite right either. Let me think about this more carefully.

Actually, the time of flight is t = √(2H/g), where H is the height of the hole above ground. The horizontal velocity is v = √(2gh), where h is the depth of the hole below the water surface. So the horizontal distance is:

d = v × t = √(2gh) × √(2H/g) = √(4ghH/g) = √(4hH) = 2√(hH)

There we go. The horizontal distance depends on the square root of the product of h and H.

The Key Insight

Here's the crucial part: h + H = constant (assuming the tank height is fixed). If the water surface is at some height and the ground is at another, then h + H = (height of water surface) - (height of ground).

When you're trying to maximize d = 2√(hH), you're essentially trying to maximize the product hH, given that h + H is fixed.

And there's a mathematical principle here: for a fixed sum, the product is maximized when the two numbers are equal. So hH is maximized when h = H.

This means the maximum range occurs when the hole is halfway between the water surface and the ground.

For more on this topic, read our article on which fraction is equivalent to 3 4 or check out johnny chan by mitch raycroft book summary.

What About Two Specific Holes?

So if you have two holes—one higher up and one lower down—the lower hole will generally shoot farther, because:

It has higher exit velocity (greater h)
It has less time to fall, but the increased velocity more than compensates

Unless... wait, let me think about this again. If one hole is much higher, it might have more time to fly, even with slower velocity.

Actually, let me reconsider the setup. If both holes are below the water surface and above the ground, then the one with greater h (deeper below surface) has greater exit velocity. But it also has greater H (higher above ground), so it has more time to fall.

The question is which effect dominates.

Let's say hole 1 is at depth h₁ below surface, height H₁ above ground. Hole 2 is at depth h₂ below surface, height H₂ above ground.

We assume h₁ + H₁ = h₂ + H₂ = constant (the total tank height from surface to ground).

Distance for hole 1: d₁ = 2√(h₁H₁) Distance for hole 2: d₂ = 2√(h₂H₂)

Since h₁ + H₁ = h₂ + H₂, and we want to compare d₁ and d₂, we can use the fact that for a fixed sum, the product is maximized when the terms are equal.

So if h₁ ≠ h₂, then one of d₁ or d₂ will be smaller than the maximum possible distance.

If h₁ > h₂ (hole 1 is deeper), then H₁ < H₂ (hole 1 is lower to ground). The product h₁H₁ could be greater or less than h₂H₂, depending on the specific values.

But here's the key insight: the maximum occurs when h = H, meaning the hole is halfway down.

So if both holes are above the halfway point (both shallow), then the lower one (larger h) will have greater range. If both holes are below the halfway point (both deep), then the higher one (smaller h) will have greater range. If one is above and one is below the halfway point, then the one closer to the halfway point will have greater range.

Wait, this is getting confusing. Let me restart with a concrete example.

A Concrete Example

Let's say the water surface is 10 meters above the ground. So the maximum possible distance occurs when h = H = 5 meters, giving d = 2√(5×5) = 2×5 = 10 meters.

Now suppose we have two holes:

Hole A: 2 meters below surface, 8 meters above ground
Hole B: 6 meters below surface, 4 meters above ground

For hole A: d_A = 2√(2×8)

For hole A: d_A = 2√(2×8) = 2√16 = 2×4 = 8 meters For hole B: d_B = 2√(6×4) = 2√24 ≈ 2×4.9 = 9.8 meters

Even though hole A is closer to the surface (shallower), hole B shoots farther because it's closer to the optimal halfway point.

Let's try another pair:

Hole C: 1 meter below surface, 9 meters above ground
Hole D: 3 meters below surface, 7 meters above ground

For hole C: d_C = 2√(1×9) = 2√9 = 2×3 = 6 meters For hole D: d_D = 2√(3×7) = 2√21 ≈ 2×4.58 = 9.16 meters

Again, the deeper hole (D) shoots farther, but notice how much closer it is to the maximum distance compared to hole C.

What about holes on opposite sides of the halfway point?

Hole E: 3 meters below surface, 7 meters above ground
Hole F: 7 meters below surface, 3 meters above ground

For hole E: d_E = 2√(3×7) = 2√21 ≈ 9.16 meters For hole F: d_F = 2√(7×3) = 2√21 ≈ 9.16 meters

Interesting! Both holes have the same range because they're equidistant from the halfway point.

The General Rule

When comparing two holes at different heights:

If both holes are on the same side of the halfway point, the one closer to halfway will shoot farther
If the holes are on opposite sides of the halfway point but equidistant from it, they'll have the same range
The maximum range is always achieved when the hole is exactly halfway between the surface and the ground

This explains why, in practice, a hole near the bottom of a tank often shoots water farther than a hole near the top – it's usually closer to that optimal halfway point, and the increased exit velocity from the greater depth more than compensates for the reduced falling time.

Conclusion

The physics of fluid jet range reveals a beautiful optimization principle: maximum distance occurs when the hole divides the vertical distance between water surface and ground equally. This simple relationship – d = 2√(hH) – governs everything from drinking straws to dam spillways. Whether you're designing a water feature or just wondering why your aquarium's leak sprays across the room, the same mathematical principle applies: position matters, and symmetry often yields the most dramatic results.

A Large Open Tank Has Two Holes