# How to Find the Depth of a Well by Dropping a Rock

TR Smith is a product designer and former teacher who uses math in her work every day.

An old trick for calculating the depth of a well or deep hole is to drop a rock into the well and measure how long it takes before you hear the rock hit the bottom. If the time between dropping the rock and hearing its thud is T seconds, then the depth of the well in meters is given by the formula

Depth = 340.29T + 11816.05 - sqrt(8041766.9T + 139619023.38) meters

For example, if you drop a rock into a well and it takes 2.9 seconds before you hear it hit the bottom, then plugging T = 2.9 into the formula above gives you a depth of 38.09 meters. It is important to release the rock from your hand at a height level with the opening of the well. Dropping it from high above the opening of the well will give you an answer that is larger than the true depth. This tutorial will show you how to derive this magical formula using the laws of physics and a little algebra.

## How the Formula is Derived

Let D be the depth of the well in meters. If you drop a rock into it, the number of seconds S for it to hit the bottom is related to D by the formula

D = 4.9S^2,

where 4.9 is half the gravitational acceleration constant, in metric units. This is the classical equation for the motion of a free-falling body in air ignoring air resistance. As soon as the rock hits the bottom of the well, the number of seconds X it takes for the sound to reach your ear at the top of the well is related to D by the formula

D = 340.29X,

where 340.29 is the speed of sound in meters per second. The speed of sound in air actually varies depending on temperature and humidity (and slightly depending on pressure), but 340.29 m/s is an acceptable typical value when you want to avoid too many variables. Therefore, the total number of seconds T from when you first drop the rock to when you hear the sound is given by

T = S + X

Since both S and X can be rewritten as functions of D, we have

T = sqrt(D/4.9) + D/340.29

Now we just need to solve this equation for D in terms of T. Working out the algebra gives us

T - D/340.29 = sqrt(D/4.9)
(T - D/340.29)^2 = D/4.9
T^2 - 2TD/340.29 + (D/340.29)^2 - D/4.9 = 0

Applying the quadratic formula to solve for D gives us

D = 340.29T + 11816.05 ± sqrt(8041766.9T + 139619023.38)

The only issue to resolve is whether it should be a "+" or a "-" in the formula. To figure out which sign is correct, let's suppose D = 0. In that case, T must equal 0 as well, therefore we need to pick the sign that makes both sides of the equation equal to 0 when T = 0. It turns out that the minus sign is the correct sign, as it gives a result that is consistent with physical reality.

## Example

You can use this trick to measure the height or depth of other things besides deep wells. Suppose you are standing at the edge of a vertical cliff whose height you don't know. To determine the height, you drop a big rock off the cliff and with a stopwatch you find that it takes 4.6 seconds for the sound of the rock's thud to reach your ears. Plugging the variable T = 4.6 into the depth formula will give you the height of the cliff.

D = 340.39*4.6 + 11816.05 - sqrt(8041766.9*4.6 + 139619023.38)
= 13381.844 - sqrt(176611151.12)
= 13381.844 - 13289.513
= 92.331 meters.

## Accuracy Issues

The apply the formula accurately, drop the rock and position your ear right at the opening of the well. Otherwise, the answer will be less accurate.

Air resistance is ignored in this formula because it makes the equations much more complicated. In physics, there is always a trade-off between simplicity and accuracy.

Terminal velocity can also affect the accuracy of the equation. Terminal velocity is the maximum speed a falling object can reach. In other words, the force of gravity on Earth does not allow a free-falling object to accelerate indefinitely. For an object like a round rock, the terminal velocity is between 150 m/s and 300 m/s. Terminal velocity is achieved after about 15 to 30 seconds of free-fall. For a typical well, this is not an issue.

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• 2 years ago

This is pretty neat. I remember doing this in high school physics using a simpler formula that didn't account for the time it takes for the sound waves to reach your ear.

• Almighty Darkseid 4 weeks ago

Does it help in any way e.g. Accruracy if I know the mass and weight of the rock/object?

Will I have to change any variables?

• Author

TR Smith 4 weeks ago from Eastern Europe

Hi AD, thanks for the question. When the only force that acts on a falling body is gravity, two objects of different masses will hit the ground at the same time. For heavy enough rocks you can safely ignore air resistance. Air resistance has a greater effect on very light objects.