Free fall distance:
how far something drops in a given time

What counts as free fall in this formula

Free fall means gravity is the only force acting on the object. It starts at rest, moves straight down, and nothing else pushes or holds it. Under those conditions every object near Earth's surface gains speed at the same rate, about 9.81 metres per second every second, regardless of its mass.

That equal-rate result surprises many people because everyday experience says a feather falls slower than a coin. The feather is slowed by air, not by gravity. In a vacuum chamber both take the same time, which astronauts demonstrated on the Moon with a hammer and a feather.

So the formula below is an idealisation. It works well for dense, compact objects over short drops, and it gets steadily worse for light, wide or fast-moving ones.

The distance formula and its companions

Distance fallen grows with the square of elapsed time. Waiting twice as long gives four times the distance, because the object is both moving for longer and moving faster.

The third version is the one you need when you know a height and want the time. It is simply the distance formula rearranged for t.

Worked example: a stone dropped for three seconds

A stone is released from rest and takes 3 seconds to reach the bottom of a shaft. Use g = 9.81 m/s² and ignore air.

The exact answer to the stated inputs is 44.145 m, but a stopwatch reading rarely deserves three decimals. If the timing is good to a tenth of a second, 3.0 ± 0.1 s gives anywhere from about 42.5 m to 45.8 m, so quote the depth as roughly 44 m.

The reverse check works too: an 80 m drop takes √(2 × 80 ÷ 9.81) = √16.31, about 4.04 s, ending at roughly 39.6 m/s.

Quick reference: distance and speed each second

Look at the middle column: 4.9, 19.6, 44.1. The gaps between successive seconds are 4.9, 14.7 and 24.5 m, which is the ratio 1 : 3 : 5. Galileo noticed that pattern long before anyone could measure g precisely.

The table also doubles as a sanity check on any app or spreadsheet. If your own figure for 4 seconds is far from 78 m, the time is probably being entered in the wrong unit.

Why real drops fall short of the ideal number

Air resistance grows with speed, so a falling object eventually stops accelerating and reaches terminal velocity. For a compact steel ball that happens at a very high speed, so the ideal formula is excellent for the first few seconds. For a skydiver in a spread position terminal speed is roughly 55 m/s, and for a raindrop only a few metres per second.

• Light, flat objects such as paper, leaves and foam lose accuracy within a fraction of a second.
• Drops longer than about 5 s from a tall structure should include drag, which needs a numerical model rather than the simple formula.
• Gravity varies slightly: about 9.78 m/s² at the equator and 9.83 m/s² near the poles, so the answer shifts by well under one percent.

The same formula on other worlds

Only g changes when you move to another body, which makes the formula a handy comparison tool. Keep the 3-second drop from earlier and swap in the local gravitational acceleration.

On the Moon the same stone would fall only 7.29 m, which is why astronauts seem to float so slowly. The ratio of the two distances is just the ratio of the g values, about 6 to 1.

A second worked check: a stone dropped from a bridge hits the water 2.5 s later. Distance is 0.5 × 9.81 × 2.5² = 30.66 m, so the deck is roughly 31 m above the water. Running it backwards, √(2 × 30.66 ÷ 9.81) returns 2.5 s, confirming the pair of formulas agree.

• Write down g with its source, because rounding to 10 m/s² adds about 2 percent.
• Report depth with the same number of significant figures as your timing deserves.
• State clearly whether you ignored air, so a reader knows how to treat the result.

Estimating drops in your head

Since g is close to 10 m/s², half of g is about 5, so the distance in metres is roughly five times the square of the seconds. Two seconds is 20 m, three seconds about 45 m, four seconds about 80 m. The estimate runs about two percent high compared with 9.81.

Counting aloud works for rough work: 'one-thousand-one' and its relatives are close to a second each. Anything under two seconds is too short for this method to be trusted, because a human's reaction time is a large fraction of the total.

For heights, the inverse is the square root of the height divided by five. A 45 m drop takes √9 = 3 s; a 20 m one takes 2 s. This mental rule is handy for sanity-checking a more careful answer.

Mistakes when timing a drop in the real world

Sound is the biggest trap when estimating the depth of a well or a canyon by dropping a stone. The splash or impact noise has to travel back up, so the time you hear is the fall time plus the sound's return time at about 343 m/s in air. Over a 44 m shaft that return takes about 0.13 s, which shifts the estimate by roughly a metre if you ignore it.

Reaction time also matters. Starting and stopping a stopwatch by hand adds typically 0.1 to 0.2 s of error, which on a 3 s fall is several metres. Filming the drop and counting frames gives better precision than a thumb on a button.