How deep is the well?

Imagine you are standing next to a well of unknown depth. You want to estimate the distance from the surface down to the water. You pick up a nearby stone and drop it into the well, carefully starting your stopwatch at the exact time you release the stone.

After 5.8 seconds, you hear a pleasing splash. How deep is the well?

(Please don’t actually drop things down wells, that’s like littering, and especially don’t do it to random holes or mineshafts you might come across. Whilst the chances are slim, you have no idea if there might be a bunch of innocent spelunkers down below who would not appreciate the added danger of missiles from above).

High school Physics equips us with simple equations of motion. Gravity (and thus acceleration) are constant for a free-falling rock. We can use the trusted equation:

If we are careful not to give the stone any initial velocity (we drop it into the well, not throw it in), then u = 0. If we drop the stone from a position parallel to the ground, there is no offset. The acceleration term is just g.

Using the value of g = 9.81 m/s/s we get an approximation that s ≈ 165m

If 165m seems like a pretty deep well, you might be interested to learn that the deepest hand-dug (as opposed to drilled) well in the World is the Woodingdean Water Well in Brighton, UK. It took four years to dig from 1858-1862, and has a depth of 390m (As deep as The Empire State building is tall! That is staggeringly deep!)

(The photo above is not the Woodingdean Well, but the 200ft deep water well at Grey’s Court in Oxfordshire, UK).

It’s not quite as simple as the calculation above.

Light travels pretty fast, but sound does not. The splash you hear when the rock hits the water is actually a pressure wave that travels through the air (the surface of the water vibrates, which oscillates the air touching it, which oscillates the air touching that …) These pressure waves travel at the speed of sound (the definition of what this means!) At 20°C the speed of sound is 343 m/s.

You hear the splash delayed by the time it takes the sound of the impact to travel back up the well to your ear. The delay is proportional to the depth of the well.

The time we record on the stopwatch is the combination of both the time the rock takes to fall plus the time it takes for the sound to reach the surface. Our simple calculation is over estimating the depth of the well because it attributes all the time to just the fall. Let’s correct that.

We’ll define the speed of sound to be c. The time for the sound to get to the surface is therefore s/c.

The true time we need to integrate over to get the distance of the drop is the time we hear the splash, subtract the time for the sound to reach us. Putting this in, collecting terms, solving the quadratic and taking the sensible root, gives the following formula for the depth of the well.

Plugging in the values: g = 9.81 m/s/s, c = 343 m/s, t = 5.8 s, we get the update estimate that s ≈ 142.3m.

Here’s a table showing the true height of a well (left column), the time the rock falls in free fall, the time for the sound to travel back, and the total time (as recorded by the stopwatch). The next column shows what the depth would suggest if we ignored the speed of sound, and the final column is the delta (how much it would we overcount).

Whilst the time the sound travels is directly proportional to depth of the well, the rock is accelerating, and this makes changes proportional to time squared. The deeper the well, the more this error grows. Here’s a graph showing the estimated depth of the well (y-axis), against the time (x-axis), for the two models.

You can see that the simple model consistently over-estimates the depth because it thinks the time for the sound to travel upwards is time that the stone is still falling down.

Let’s go further down the rabbit hole. The rock is not falling in a vacuum, but instead is flowing through air. Air is a dense fluid and imparts drag on the stone. At slow speeds it’s appropriate to ignore drag, but as speed increases, we don’t have that luxury (wave your hand in front of your face, and you can’t feel any resistance, but stick you hand out of the window of a car travelling at highway speeds and you certainly can!).

Drag is proportional to the square of the speed.

Here ρ is the density of the air, S (capital S in this case) is the wetted area, v is the velocity, and C_D is the drag coefficient (aerodynamically smooth shapes have lower drag coefficients, those with more obtuse shapes have higher ones).

As the rock accelerates down the shaft, and its speed increases, then it feels more and more resistance with the air. The drag opposes the pull of gravity, decreasing the net downward force.

From Newton II, we get the following relationship between the forces and the acceleration:

It’s worth taking a slight detour at this point to talk about the concept of terminal velocity, both because it is interesting, but also because it will allow us to simplify the math a little later.

As the velocity of a freefalling object in a fluid increases, so does its drag (proportional to speed squared). Eventually this drag force increases until it equals the force exerted by gravity. When this happens, there is no net force acting on the body and it stops accelerating. Outside of any external force, it’s not going to get any faster. It has reached it’s terminal velocity. We will denote this speed as v_t.

We can see from this equation that if the density of the air is lower (high altitude), the terminal velocity is higher, and that the lower the drag coefficient, the higher the terminal velocity. More massive objects also have higher terminal velocities.

Sky divers, jumping out of planes, rapidly reach their terminal velocity when in free fall. By modifying their pose (shape), they can change their drag coefficient (and cross sectional area) and regulate their speed and thus their rate of descent: Laying out spread eagle, (arms and legs out wide) produces a high drag coefficient, but by pinning their arms to their side, and lowering their head, they get aerodynamically smoother, and speed up. A skydiver in a stable belly flat position has a terminal velocity of around 120 mph (just over 50 m/s). Lowering a head and tucking in can increase this speed to around 180 mph.

I’ve never attempted skydiving; it’s not something that has appeal to me, so I’m still a member of the set of people whose lifetime total number of takeoffs == total number of landings (sometimes off by one, if I’m in the middle of a trip!), but I’ve been told that in freefall you don’t feel like you are falling (and why would you?, there are no net forces acting on you; it should feel no different than laying on a bed).

When a skydiver pops their chute, there is a massive increase in drag, and this slows down the terminal velocity sufficiently enough to provide a safe landing speed.

A baseball and a golf ball both have a terminal velocities of around 33 m/s (75 mph)*. Small hailstones are at 14 m/s (30 mph), and raindrops at 9 m/s (20 mph).

The relationship between mass, area, and drag, to terminal velocity explains why small rodents can survive tremendous falls pretty much unharmed. There’s a wonderful quote from the paper On Being the Right Size by the great scientist J.B.S. Haldane (Haldine united Darwin's theory of evolution with Mendel's laws of inheritance, and was one of the most colorful figures in 20th-century genetics; an accomplished statistician and scientist).

(As exciting as a horse splash sounds, please do not throw any living object down a mine shaft either!)

Ignoring for a moment the sound delay (we know how to account for that), and concentrating only on the drop part of the timing we have the equation:

Separating variables, we get

Re-arranging the formula for terminal velocity, and substituting in

We get

Referencing our table of standard integrals, we remember that

We get the solution that:

This is an interesting result that shows that, mathematically, the terminal velocity is never quite reached. We just continue to get closer, and closer, and closer (very quickly, imeasurably close). This is what we’d expect for a purely mathematical perspective.

To get the distance fallen we integrate one more time (again referencing our table of standard integrals and remembering tanh(x)=sinh(x)/cosh(x) ).

Putting it all together and adding back in the time delays for the speed of sound we get:

Which is just a little too gnarly to solve for s, so instead we’ll just solve numerically. Let’s also use some real-world figures.

Let’s imagine our rock is a marble-sized spherical piece of granite with a diameter of 2 cm. The density of granite is around 2.75 g/cm³, so this rock will weigh approx. 11.5 g = 0.0115 Kg. It’s cross-sectional area is 3.14 cm² = 0.000314 m². At 20°C, the density of air is 1.2 Kg/m³, and the rough approximation for the drag coefficient for a sphere is C_D = 0.4

Plugging these in gives a projected terminal velocity for the stone as approx. 38.7 m/s (87.1 mph)

Here’s a plot of the displacement of the stone against time (also plotted on the same graph is the simple approximation based on the equations of motion). You can see the curve that the takes into account air resistance flattens out and almost becomes a straight line is it approaches terminal velocity (and thus constant speed). Because of aerodynamic drag, and it’s effect on the relatively light weight rock, the distance it travels in 5.8 seconds is just 118.6 m, and not the 165 m predicted by the simple model (and this is before we put in the modification for the time delay).

If, instead of using a pebble, we dropped a cast iron cannonball down the well*, then with a radius of 10 cm and density of 7.13 g/cm³, it will have mass of 29.9 Kg, giving a terminal velocity of 197 m/s (441 mph). This terminal velocity is so high that the drag is negligible compared over such a short distance, so the curves are pretty similar.

If we dropped a feather down, we’d probably have time to make a cup of tea before it hit the water!

*Do I need to remind you that it’s not a good idea to throw cannonballs down wells?

Below is a final plot showing the relationship between the height of well, and time it would take to hear the splash taking into account quadratic wind resistance, and the propagation delay of the sound come up from the bottom for a small spherical pebble of granite (I’ve swapped the axes over on this chart).

Plotted on the same graph is the simple approximation the basic equations of motion give. It’s a pretty big difference.

Using numeric methods (I just used Excel) on the full model gives an estimte of a depth of 114.9m

Of the 5.8s recorded on the stopwatch, this works out to a fall time of 5.46s and a sound trip return of 0.34s. Interestingly, because the terminal velocity limits the speed the stone can fall, this reduces the error caused by not applying the time delay.

Taking this to the limit, if we dropped our pebble down the 390m Woodingdean water well, we would not expect to hear a splash until 12.8 seconds had elapsed (and 1.13 seconds of this would have been the time it took for the sound to travel back up the well).

Try counting that out in your head. That really is a deep well!