Free Fall Time Calculator

Enter drop height h. An empty g field means 9.81 m/s². The calculator finds the time of a fall from rest: t = √(2h/g). From 10 m the result is about 1.43 s; from 100 m about 4.52 s. The panel also shows v = g·t just before the ground. The model assumes v₀ = 0 and no air resistance — it is not a downward throw and not height-from-time as the primary input.

Inputs

Result

Enter the height and, optionally, g. The fall time, impact speed, and the Galileo ladder will appear here.

Know the time and need the height? Use h = ½gt² in the free fall distance calculator. The value of g for a planet comes from the g = GM/r² calculator.

How it works

How t is computed for 10 m and g = 9.81 m/s²
2h20 m
÷ g9.81 m/s²
√2.039 s²
t1.43 s
h t = ? t = √(2h/g)
An object released from height h falls for t = √(2h/g). Time grows with the square root of the height, not in proportion to it.

Free fall is the motion of a body acted on by gravity alone. When an object is released from rest, its distance grows with the square of time: h = ½gt². Inverting that relation gives the fall time: t = √(2h/g), where h is the drop height and g the gravitational acceleration. This is Galileo’s kinematics of uniformly accelerated motion, standard from the Discorsi of 1638 to today’s general physics courses (Halliday, Resnick, Walker; Young, Freedman).

The calculator works with two numbers: the height h and the acceleration g; an empty g field means 9.81 m/s². It first evaluates the quantity under the root, 2h/g, and then takes its square root. For 10 m you get 2·10/9.81 = 2.039 s², whose square root is t = 1.43 s. For a 0.75 m countertop the radicand is 0.153 s² and the time 0.391 s.

What the result panel shows. Besides the fall time, the calculator reports the speed just before the ground, v = g·t (about 14.0 m/s or 31 mph for 10 m), the average speed over the whole drop (half the final speed), the number under the root, the time to the halfway point (t/√2, about 1.01 s for 10 m), and the entered h and g. The Galileo ladder under the result shows what fraction of the height the object has covered after one quarter, one half, and three quarters of the time: 1/16, 1/4, and 9/16. The chart compares fall times from 1, 2, 5, 10, 20, and 50 m at the entered g.

Limits of the model. The formula assumes zero initial speed, constant g along the whole path, and no air resistance. The calculator does not handle a downward or upward throw (that is the vertical throw calculator), terminal velocity, or the shape of the object. For a compact body dropped from a few to a few tens of meters the error is small; for a sheet of paper, a leaf, or a skydiver the real time is clearly longer.

The unit switch in the header changes the fields to feet and ft/s² and the speeds to ft/s and mph; the values in the text stay in SI units as a reference. A comma and a period in the fields mean the same decimal separator. The height must be greater than zero: there is no fall from zero height, so t = 0 is not a result worth showing.

Examples on this page: 0.75 m → 0.391 s; 2 m → 0.639 s; 10 m → 1.43 s; 25 m → 2.26 s; 114 m → 4.82 s; 2 m on the Moon at g = 1.62 m/s² → 1.57 s. The “Fill this example” button enters the same numbers into the fields and refreshes the result panel, the ladder, and the chart.

Formula

t = √(2h / g)

[t] = √(m / (m/s²)) = √(s²) = s; v = g·t, t½h = t/√2, vavg = v/2

Model: uniformly accelerated motion from rest (Galileo, Discorsi 1638; in Newtonian notation h = ½gt²). The rearrangements h = ½gt² and g = 2h/t² follow from the same formula, but this calculator solves only for t. Order of computation: 2h/g in SI units, square root, then v = g·t and t/√2.

Theory, derivation, and domain

Definition of free fall and the time formula

Free fall is the motion of a body acted on by gravity alone; near the Earth’s surface that force gives every body the same acceleration g, independent of mass. A body released from rest moves with uniform acceleration: its speed grows linearly, v = g·t, and its distance quadratically, h = ½gt². The time to fall a height h is therefore the square root t = √(2h/g).

Two properties follow from the form of the formula, and both clash with everyday intuition. First, time grows with the square root of height: a building four times taller gives a fall only twice as long. Second, time does not depend on mass — in a vacuum a lead ball and a feather take the same time. On Earth the differences between objects come from air resistance, which this model ignores.

Historical note: Aristotle, Galileo, and the Apollo 15 hammer

For almost two thousand years Aristotle’s view prevailed: heavier bodies fall faster, with speed proportional to weight. Galileo Galilei undermined it first with a thought experiment (two bodies tied together would have to fall both faster and slower than each alone) and then with measurements. Because a vertical drop was too fast for the clocks of his day, Galileo slowed the motion by rolling balls down an inclined plane and timing them with a water clock.

In Discorsi e dimostrazioni matematiche intorno a due nuove scienze (1638) he stated the law: distances covered in successive equal intervals of time stand in the ratio of the odd numbers 1 : 3 : 5 : 7, which is equivalent to h ∝ t². The most famous demonstration of the principle remains David Scott’s experiment on Apollo 15 (1971): a hammer and a falcon feather released together on the airless Moon struck the ground at the same instant.

Deriving t = √(2h/g) from the equations of motion

Step 1. Take an axis pointing downward with its origin at the release point and initial speed v₀ = 0. The acceleration is constant and equal to g, so the speed at time t is v(t) = g·t. Step 2. Distance is the integral of speed over time: h(t) = ∫₀ᵗ g·τ dτ = ½gt². The same result is the area of the triangle under the v(t) graph with base t and height g·t.

Step 3. Solve h = ½gt² for time: t² = 2h/g, and since time is non-negative, t = √(2h/g). Step 4. The speed just before the ground follows from v = g·t = √(2gh); energy conservation, mgh = ½mv², gives the same value with the mass canceling. For h = 10 m and g = 9.81 m/s²: t = √(2.039) = 1.43 s, v = 14.0 m/s. The time to the halfway point is t/√2 ≈ 0.707·t, because h/2 = ½g(t/√2)².

Model assumptions: rest, constant g, no drag

The formula t = √(2h/g) is exact only under three assumptions. First: zero initial speed, meaning the object is released, not thrown. A downward throw shortens the time and an upward throw lengthens the total flight; both cases are covered by the more general equation h = v₀t + ½gt², handled by a separate vertical throw calculator. Second: the acceleration g is constant along the path. For heights of meters to kilometers this is an excellent approximation, because g decreases by only about 0.3 % per 10 km of altitude.

The third and, in practice, most important assumption is the absence of air resistance. Drag grows with speed, so the longer the fall, the more the real motion departs from the model. For a compact, dense object — a stone, a key, a ball — dropped from a few to a few tens of meters, the time differs by no more than a few percent. For light, extended objects (a sheet of paper, a leaf, a balloon) the model breaks down within a fraction of a second.

Air resistance and terminal velocity

At speeds typical of a fall, the drag force is approximately proportional to the square of the speed: F = ½·C·ρ·A·v², where C is a shape coefficient, ρ the air density (about 1.2 kg/m³), and A the cross-sectional area. When drag equals the weight m·g, the acceleration drops to zero and the body continues at a constant terminal velocity vt = √(2mg/(C·ρ·A)).

For a skydiver in a belly-down position the terminal velocity is about 50–55 m/s (110–120 mph), reached after roughly 12 seconds and 450 m of fall; the vacuum model would predict 118 m/s by then. A 2 mm raindrop reaches only about 6.5 m/s. On the other hand, a steel ball dropped from 10 m hits the ground at a speed only slightly below the 14 m/s the formula predicts. The practical rule for this page: the larger the ratio of mass to cross-sectional area and the lower the height, the better the result describes reality.

The value of g: Earth, other worlds, and measurement

The standard acceleration of gravity gn = 9.80665 m/s² was adopted by the 3rd General Conference on Weights and Measures in 1901 as a conventional value. The real g at the Earth’s surface ranges from about 9.780 m/s² at the equator to 9.832 m/s² at the poles, mainly because of the planet’s flattening and the centrifugal effect of its rotation; across the contiguous United States it is close to 9.80 m/s², so the default of 9.81 changes a 10 m fall time by about a thousandth of a second. Local deviations of the order of 10⁻⁵ m/s² are measured with gravimeters and used in geodesy and mineral exploration.

On other bodies of the Solar System g is different, and with it the fall time, which scales as 1/√g. On the Moon (1.62 m/s²) a 2 m drop takes 1.57 s instead of 0.639 s on Earth — hence the slow-motion look of Apollo footage. On Mars (3.71 m/s²) the same drop takes 1.04 s, and on Jupiter, if it had a surface (24.79 m/s²), only 0.40 s. The g for any mass and radius comes from the g = GM/r² calculator; the result can be entered into the g field on this page.

Sources

The definition, derivation, and historical note come from general physics textbooks and metrology sources: G. Galilei, Discorsi e dimostrazioni matematiche intorno a due nuove scienze, Leiden 1638 (English: Dialogues Concerning Two New Sciences, Crew and de Salvio translation); D. Halliday, R. Resnick, J. Walker, Fundamentals of Physics, chapter on motion along a straight line; H. D. Young, R. A. Freedman, University Physics, chapter “Motion Along a Straight Line”; R. P. Feynman, R. B. Leighton, M. Sands, The Feynman Lectures on Physics, vol. I, chapters 8 (Motion) and 9 (Newton’s Laws of Dynamics); BIPM, The International System of Units (SI), 9th edition, and the 3rd CGPM resolution (1901) on gn; NASA, Apollo 15 Lunar Surface Journal, the hammer-and-feather demonstration. This page is for educational purposes.

How to use

  1. Enter the drop height h in meters, for example 10 for a diving platform or 0.75 for a countertop.
  2. Leave the g field empty for Earth (9.81 m/s²). For the Moon enter 1.62, for Mars 3.71.
  3. Click Calculate or wait for the live result. For 10 m and 9.81 m/s² the calculator shows 1.43 s.
  4. Check the panel: speed at the ground, average speed, the number under the root, the time to half the height, and the Galileo ladder with its 1 : 3 : 5 : 7 gaps.
  5. The unit switch in the header changes the fields to feet and ft/s² and the speeds to ft/s and mph. The Clear button removes the data and hides the panel.

Symbol glossary

t is the fall time, h the drop height, g the gravitational acceleration. For 10 m and 9.81 m/s² the result is 1.43 s.

t
Fall time [s] from rest to the ground, t = √(2h/g).
h
Height [m] from which the body is released.
g
Gravitational acceleration [m/s²]; an empty field means 9.81.
2h/g
The number under the root [s²]; for 10 m and 9.81 m/s² it equals 2.039 s².
v
Speed just before the ground [m/s], v = g·t = √(2gh); the average over the drop is v/2.
t/√2
Time to half the height, about 0.707·t; the second half of the distance takes only 0.293·t.
v₀
Initial speed; always zero in this model (a release, not a throw).

Real-life examples

Example 1

A mug slides off a countertop at {{u:0.75:dist-m:2}}. Human reaction time is about 0.2 s, so less than two tenths of a second remain for the hand to move.

t = 0.391 s

Example 2

A book falls from the top shelf at {{u:2:dist-m:0}}. At the floor it is moving at about {{u:6.26:speed-ms:1}}.

t = 0.639 s

Example 3

A diver steps off a {{u:3:dist-m:0}} pool platform with no upward bounce. The result is the time from leaving the board to reaching the water.

t = 0.782 s

Example 4

An Olympic diving platform is {{u:10:dist-m:0}} high. That is all the time a diver has for every rotation before entering the water at about {{u:14:speed-ms:0}}.

t = 1.43 s

Example 5

A stone released from a bridge {{u:25:dist-m:0}} above a valley. Just before the water it moves at about {{u:22.1:speed-ms:1}}, close to 50 mph.

t = 2.26 s

Example 6

An object falls from an observation deck at {{u:114:dist-m:0}}. This is a purely arithmetical example: in reality air resistance would lengthen this time.

t = 4.82 s

Example 7

A pebble knocked off the edge of a {{u:45:dist-m:0}} cliff. Three seconds of silence before the impact is a good test of the formula.

t = 3.03 s

Example 8

A stone drops into a well {{u:15:dist-m:0}} deep. The splash is heard after the fall time plus a few hundredths of a second for the sound to return.

t = 1.75 s

Example 9

An object falls from the roof of a {{u:55:dist-m:0}} tower. The impact speed would exceed 70 mph if there were no air resistance.

t = 3.35 s

Example 10

An astronaut drops a soil sample from {{u:2:dist-m:0}} at the lunar g = {{u:1.62:accel:2}}. The fall lasts two and a half times longer than on Earth.

t = 1.57 s

Example 11

A window box falls from the fourth floor, about {{u:12:dist-m:0}} above the sidewalk. One and a half seconds is not enough for anyone to step aside.

t = 1.56 s

Example 12

The same {{u:2:dist-m:0}} drop, but on Mars at g = {{u:3.71:accel:2}}. Time scales as 1/√g.

t = 1.04 s

Frequently asked questions

What does this calculator compute, and from which formula?

The calculator finds the fall time of a body released from rest at height h using t = √(2h/g). It first evaluates the number under the root, 2h/g, and then takes its square root. For 10 m and g = 9.81 m/s² the result is 1.43 s.

Which units should I use for the height and g?

By default the height is in meters and g in meters per second squared; the result is in seconds. After switching the header to US units the fields take feet and ft/s², and speeds are shown in ft/s and mph. A comma and a period mean the same decimal separator.

What happens if I leave the g field empty?

The calculator uses 9.81 m/s², a typical value of the Earth’s gravitational acceleration. The standard value 9.80665 m/s² differs from it by less than 0.1 %, which changes a 10 m fall time by about a thousandth of a second.

Why does the time grow so slowly with height?

Because the height sits under a square root. Four times the height gives twice the time: a 2 m drop takes 0.639 s and an 8 m drop 1.28 s. One hundred times the height lengthens the time only tenfold.

Does the mass of the object affect the fall time?

Not in the free fall model: the acceleration g is the same for all bodies, so the mass cancels in the equations. In a vacuum a hammer and a feather fall together, as the Apollo 15 demonstration showed. On Earth the differences between objects come only from air resistance.

How fast is the object just before it hits the ground?

The result panel reports v = g·t, equivalently v = √(2gh). For 10 m it is about 14.0 m/s, or 31 mph; for 25 m about 22.1 m/s, close to 50 mph. The average speed over the whole drop is half of that.

What does the Galileo ladder under the result show?

The four markers correspond to one quarter, one half, three quarters, and all of the fall time. The distance covered grows as 1 : 4 : 9 : 16, so the successive gaps stand in the ratio 1 : 3 : 5 : 7. After half the time the body has covered only one quarter of the height.

Does the formula work for an object that is thrown rather than released?

No. The formula t = √(2h/g) assumes zero initial speed. A downward throw shortens the time and an upward throw lengthens the total flight. Both cases are described by h = v₀t + ½gt², which a separate vertical throw calculator handles.

How much does air resistance change the result?

For a compact, dense object dropped from a few to a few tens of meters the difference is no more than a few percent. Above roughly 50 m, and for light or extended objects, the real time is clearly longer because the body approaches terminal velocity. The calculator shows the vacuum result and says so in the note under the result.

How do I compute the fall time on the Moon or Mars?

Enter the matching g: Moon 1.62 m/s², Mars 3.71 m/s². Time scales as 1/√g, so a 2 m drop takes 1.57 s on the Moon, 1.04 s on Mars, and 0.639 s on Earth. The g for another celestial body comes from the g = GM/r² calculator.

I know the fall time. How do I find the height?

This calculator solves only for t. The inverse problem, h = ½gt², is handled by the free fall distance calculator: for 2.5 s and g = 9.81 m/s² the height is about 30.7 m. A stopwatch time also contains the reaction time, usually 0.15–0.25 s.

Why is there no result for a height of zero or a negative height?

There is no fall from zero height, and a negative height has no meaning in this model. The calculator requires h > 0 and g > 0; as long as either value is out of range it shows a message instead of a result.

How do I calculate free fall time?

For a drop from rest: t = √(2h/g). At h = 10 m and g = 9.81 m/s² you get about 1.43 s. Mass does not appear in this model.

How long does a fall from 100 m take?

About 4.52 s at g = 9.81 m/s². √(200/9.81) ≈ 4.52. Impact speed is then about 44 m/s — with no air drag.

How long does a fall from 50 m take?

About 3.19 s. √(100/9.81) ≈ 3.19. Time grows with the square root of height, not linearly.

Does this solve for height from fall time?

Not as the primary result. The input is height (and optional g); the result is time. The inverse h = ½gt² lives on the free-fall page.

How long does a fall from 100 ft take?

About 2.49 s at Earth g, because 100 ft ≈ 30.5 m. In US mode you enter height in feet; the formula stays t = √(2h/g).

The result is the square root t = √(2h/g): for 10 m and g = 9.81 m/s² it equals 1.43 s, with a speed at the ground of 14.0 m/s (31 mph). For height from time, use the free fall distance calculator.

Knowledge sources

The definition, unit, history, and derivation below come from recognized physics and metrology sources.

Page updated November 3, 2026.