Free-Fall Calculator

drop time, impact speed, and energy for free fall under gravity

Worked example — Free-Fall Calculator

Fall time (Impact)	4.5152 s
Impact speed	44.29 m/s (159.5 km/h / 99.1 mph)
Kinetic energy per kg	981 J/kg

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Free-Fall Calculator

Free fall is motion under gravity alone, with the air taken out of the problem. One constant acceleration g acts for the whole drop, and everything the panel reports follows from that single assumption — no forces are added, and none are left out.

What each symbol means

h: the drop height, measured straight down. It is the only distance in the problem: how far the object travels sideways, and what shape it has, do not appear.
g: the gravitational acceleration, in metres per second squared. Set it to 1.62 for the Moon and the same 100 m drop stretches from 4.5152 s to 11.1111 s, arriving at 18 m/s instead of 44.2945 m/s.
v: the impact speed at the moment it lands, not an average. The object was travelling slower than this for the entire fall except its final instant.
t: the fall time. It grows as the square root of the height, which is the part that surprises people — ten times the drop buys only 3.1623 times the time.

Where the formula comes from

With constant acceleration and a standing start, distance is h = ½gt². That is the whole of the physics; the rest is algebra.
Solve it for the time: t = √(2h/g). The square root is the reason a fall does not take twice as long from twice the height.
Speed after time t is v = gt. Substituting the time from step 2 gives v = g·√(2h/g) = √(2gh) — the height and the gravity enter symmetrically, and the mass never appears in either line.
Kinetic energy per kilogram is ½v², and squaring √(2gh) gives back exactly g·h. At h = 100 and g = 9.81 the third row reads 981 J/kg, which is 9.81 × 100 — the readout is the product of the two inputs and nothing else.

How to read what you see

Change the height and watch the two upper rows move together but not in step: at 1000 m the fall takes 14.2784 s against 4.5152 s, and lands at 140.0714 m/s against 44.2945 m/s — both multiplied by 3.1623, because ten times the height is √10 times everything. Quarter the height instead and the speed exactly halves: 25 m gives 22.1472 m/s.

Assumes: No air, a standing start, and a gravity that does not change over the drop. The last one is quietly the strongest: g falls off with altitude, so a genuine 1000 m drop does not run at a single value the way this one does.
Breaks when: Air resistance is what breaks it first, and it breaks it completely rather than slightly. A real skydiver stops accelerating at a terminal speed near 55 m/s, so this page already overstates the 1000 m case by more than a factor of two at 140.0714 m/s. The formula has no term that could ever produce a ceiling — v = √(2gh) rises without limit — so the number it prints is not a slightly optimistic answer but an answer to a different question.

Galileo could not time a falling ball, so he slowed gravity down 🖖

A drop of a few metres is over in under a second, and the best clock available in 1600 was a vessel of water weighed on a balance. So Galileo diluted gravity instead of fighting it: roll a ball down a shallow ramp and the same steady acceleration acts at a fraction of its strength, stretching the motion out until it can be measured by hand. What he found was that the distances covered in successive equal intervals went as 1, 3, 5, 7 — the odd numbers, whose running totals are 1, 4, 9, 16, the squares. That is h ∝ t² established without a stopwatch. The gravity box above does the same thing in software: setting it to 1.62 changes no physics, only the pace at which the physics plays out.

Practice

Check yourself

Predict the answer first, then use the controls above to find out. Reveal only after you have committed to a guess — that is what makes it practice.

Press 100 m tower (Earth): 4.5152 s, 44.2945 m/s, 981 J/kg. Now set gravity to the Moon's 1.62 and leave the height alone — the impact speed collapses to 18 m/s. How high would the Moon drop have to be to land just as hard, and which row tells you when you have got there?

Show answer

About 605.5 m, six times the height for one sixth of the gravity. The row to watch is the third one: kinetic energy per kilogram is g·h and nothing else, so matching 981 J/kg is the same as matching the speed. At 605.5 and 1.62 it reads 980.91 J/kg and 44.2924 m/s, against Earth's 981 and 44.2945. The fall now takes 27.341 s instead of 4.5152 s — six times as long, arriving at the same speed, because weak gravity needs distance to do what strong gravity does quickly.
A real skydiver stops speeding up at around 55 m/s. Above what drop height does this page start reporting a speed that no falling person reaches?

Show answer

Between 154 m and 155 m: 154 gives 54.968 m/s and 155 gives 55.1462 m/s. That threshold is lower than most people expect, and everything above it is an answer about vacuum rather than about air. The gap then widens fast — by 1000 m the page reports 140.0714 m/s, two and a half times a speed a falling body in air cannot pass at all.

Problem solved in full

Drop a stone from 100 m with g = 9.81 m/s² 5 steps
Drop a stone from 100 m with g = 9.81 m/s². Find the fall time, the impact speed and the energy per kilogram — then find where in the fall the halfway points are, because neither of them is at half the time.
1. Displacement is quadratic in time, so setting the height to zero and solving is the whole of the first step.
  
  $y (t) = h - \frac{1}{2} g t^{2} = 0 ⟹ t = \frac{2 h}{g} = \frac{200}{9.81} = 4.5152 s$
2. Speed is the derivative, evaluated at that instant. It is worth converting: 159 km/h is motorway speed, from ten storeys.
  
  $v = g t = 9.81 \times 4.5152 = 44.29 m/s = 159.5 km/h$
3. Substituting v² = 2gh into the kinetic energy makes the mass and the speed both disappear, leaving the drop height and gravity. This is potential energy handed straight back, which is why the number is a round 981.
  
  $v^{2} = 2 g h ⟹ \frac{E _{k}}{m} = \frac{1}{2} v^{2} = g h = 9.81 \times 100 = 981 J/kg$
4. Now the shape of the fall. Writing the height as a fraction of the total makes the time dependence a plain square, and squares are unkind: half the elapsed time buys a quarter of the distance.
  
  $\frac{y}{h} = 1 - (\frac{t}{t _{imp}})^{2} : t = \frac{1}{2} t_{imp} \Rightarrow only 25% fallen; y = \frac{1}{2} h \Rightarrow t = 0.707 t_{imp}$
5. Air changes the destination, not just the pace. With a terminal velocity of 55 m/s the closed form gives 38.0 m/s — 14% off the speed but 26% off the energy, because energy goes as the square, and this is the one line where the mass would come back.
  
  $v_{t} = 55 m/s : v_{imp} = v_{t} 1 - e^{- 2 g h / v_{t}^{2}} = 55 1 - e^{- 0.6486} = 38.0 m/s, \frac{E _{k}}{m} = 721.8$
Answer

4.5152 s, 44.29 m/s, and exactly 981 J/kg. That last figure is gh with nothing else in it: the mass cancels twice over, once in the weight and once in the inertia, so a pebble and a piano arrive with the same energy per kilogram. And the fall is grossly back-loaded — halfway through the time the stone has covered a quarter of the drop, and it only reaches the halfway height at 71% of the time. Half the journey happens in the last 29% of it, which is why a fall that looks survivable from the top is not.

$t = 4.5152 s, v = 44.29 m/s, E_{k} / m = g h = 981 J/kg$

Learning path

Energy instead of forces

Leads to Roller coaster the speed a drop of height h produces, v = √(2gh), and the fact that the mass cancels out of it entirely.

References (3)

Where “distance goes as the square of the time” comes from — the result the whole panel rests on: Galileo Galilei, Two New Sciences, including Centers of Gravity & Force of Percussion, translated by Stillman Drake. University of Wisconsin Press, 1974. ISBN 0-299-06404-2. First published as the Discorsi in 1638.
The 9.80665 m/s² the gravity field is built around, and where that exact figure was fixed: Bureau International des Poids et Mesures, Declaration 2 of the 3rd CGPM (1901): “The value adopted in the International Service of Weights and Measures for the standard acceleration due to gravity is 980.665 cm/s².”
The “no air” assumption tested where there genuinely is none: NASA, “The Apollo 15 Hammer-Feather Drop.” Commander David Scott released a 1.32 kg geological hammer and a 0.03 kg falcon feather from about 1.6 m; both struck the surface together.

By the 42Math team · Updated 2026-05-21

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

100 m tower (Earth) - 100 m at 9.81 m/s² → 4.5152 s and 44.2945 m/s, and the 981 J/kg of kinetic energy is g·h to the digit
100 m drop (Moon) - The same 100 m at the Moon's 1.62 m/s² takes 11.1111 s and arrives at a round 18 m/s
1000 m (Earth) - Ten times the height gives 14.2784 s, not ten times 4.5152: fall time goes as √h, so tenfold height costs √10