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বাং

Free fall: drop a coin and a feather and watch

Free fall is the motion of an object moving under gravity alone, with no air resistance or other force acting on it. Near Earth every freely falling object speeds up by g = 9.8 m/s² each second, whatever its mass, so v = gt, h = ½gt² and the time to fall is t = √(2h/g).

With no air, the coin and the feather land together

CoinFeatherNo-air predictionvPositions at equal time steps
Speed

Controls

20 m

Acceleration due to gravity, g (m/s²)

Readings

Time, t
0.00s
Formula time, √(2h/g)
2.02s
Coin velocity, v
0.0m/s
Feather velocity
0.0m/s
Coin distance fallen
0.0m
Feather distance fallen
0.0m
Coin landed at
…s
Feather landed at
…s

How to use this simulation

  1. Just watch first: air is off, so the coin and the feather fall side by side and land at the same instant.
  2. Tick “Turn air resistance on”: the coin falls almost exactly as before, but the feather drifts down slowly.
  3. Keep the strobe on: the gaps between snapshots taken at equal time steps grow as 1 : 3 : 5 : 7, and the ratios are printed beside the coin.
  4. Watch the two graphs on the right: v–t is a straight line and h–t is a curve (a parabola).
  5. Change the height and the world, then compare “Formula time” with “Coin landed at” in the readings panel.

A mango, a raindrop and a dropped phone

A ripe mango lets go of its branch. For the first instant it barely seems to move, but by the time it reaches the ground it hits hard enough to dent the grass. Its speed was growing the whole way down. By how much, exactly, and how far did it fall in each second? Answering that is what free fall is about.

Everyone knows a phone that slips off a table might survive, while one dropped from a second-floor balcony probably will not. But if you double the height, do you double the time the fall takes? Surprisingly, no. Doubling the height makes the fall only about 1.4 times longer. By the end of this page you will be able to show why with one line of algebra.

Then there is the oldest puzzle of all: drop a heavy stone and a light sheet of paper together, and which lands first? Most people say the stone, and for two thousand years so did the greatest thinkers. The real answer is “it depends on the air”. Switch the air off in the simulation above and the coin and the feather fall together.

Starting from zero: what free fall really means

Earth pulls everything towards its centre; we call that pull gravity, and on an object of mass m it is the weight mg. If you release an object and nothing else acts on it (no air pushing back, no string, no hand), gravity alone decides its motion. That is free fall.

In school problems free fall usually comes with two conditions: the object starts from rest (initial velocity u = 0), and air resistance is ignored. Air is always there in real life, but for small dense objects such as stones, coins and balls it is so weak over a few metres that ignoring it changes the answer very little.

Gravity increases a falling object’s velocity by the same amount every second. Near Earth’s surface that amount is about 9.8 metres per second, every second. This is the acceleration due to gravity, written g. After 1 s of falling the velocity is 9.8 m/s, after 2 s twice that, after 3 s three times that. Velocity grows at a steady rate, so free fall is uniformly accelerated motion.

That phrase matters. Because the acceleration never changes, the three equations of motion you learned for constant acceleration (v = u + at and the rest) apply to free fall exactly as they are. Replace a with g and s with h and you have every free fall formula. Free fall is not new physics; it is the most familiar example of constant acceleration.

Here is the beautiful part: g does not depend on mass. Earth pulls a heavy object harder, but a heavy object also needs more force to get moving, because it has more inertia. The two grow in exactly the same proportion and cancel. So without air an elephant and an ant fall with the same acceleration.

Key terms in free fall

Get the vocabulary straight before the formulas; definition questions in exams come straight from this table.

TermSymbolWhat it meansSI unit
Weight (force of gravity)mgThe pull of Earth on an object, towards Earth’s centrenewton (N)
Acceleration due to gravitygHow much a falling object’s velocity grows each secondm/s²
Initial velocityuVelocity at the start; u = 0 when an object is simply droppedm/s
Final velocityvVelocity after time t, or on reaching the groundm/s
Height, or distance fallenhHow far the object has droppedmetre (m)
TimetTime since releasesecond (s)
Air resistance (drag)—The force air exerts against the motion; it grows with speednewton (N)
Terminal velocityvₜThe speed at which drag equals weight, so the velocity stops growingm/s

Who discovered it: Aristotle versus Galileo

About 2,300 years ago the Greek philosopher Aristotle taught that heavy objects fall faster than light ones, in proportion to their weight. It sounds sensible and matches everyday experience (a stone falls faster than a leaf), so for nearly two thousand years almost nobody thought it needed testing.

In the late sixteenth century the Italian scientist Galileo Galilei (1564–1642) challenged it. The famous story says he dropped two balls of different mass from the Leaning Tower of Pisa; historians doubt it happened. What is certain is that he rolled balls down smooth inclined ramps and timed them carefully. On a gentle slope a ball moves slowly enough to time even with a water clock.

From those experiments Galileo concluded that without resistance all objects fall at the same rate, that the velocity grows in proportion to time, and that the distance grows in proportion to the square of time. These three statements are still taught as the laws of falling bodies, which is why the answer to “who discovered free fall?” is Galileo.

He also gave a brilliant thought experiment. Suppose heavy stones really do fall faster than light ones, and tie a heavy stone to a light one. The light stone should hold the heavy one back, so the pair falls slower than the heavy stone alone. But the pair is heavier than either stone, so it should fall faster. It cannot do both, so the assumption was wrong.

The three laws of falling bodies

All three laws share one condition: the object starts from rest and falls without resistance. When you write them in an exam, state that condition first.

First law: every object covers equal distances in equal times

Objects released from rest at the same height, falling without resistance, cover equal distances in equal times.

In plain words: heavy or light, big or small, without air they all hit the ground together. With air off in the simulation, the “landed at” times of the coin and the feather are identical, because g does not depend on mass.

Second law: velocity is proportional to time

The velocity gained by a body falling freely from rest is proportional to the time of fall.

Double the time and you double the velocity. It follows from v = u + gt: with u = 0, v = gt, and since g is constant, v ∝ t. That is why the v–t graph is a straight line through the origin whose slope is g = 9.8.

v ∝ tv₁ / v₂ = t₁ / t₂

Third law: distance is proportional to the square of time

The distance fallen by a body falling freely from rest is proportional to the square of the time of fall.

Double the time and the distance becomes four times as large; triple it and it becomes nine times as large. It follows from h = ut + ½gt²: with u = 0, h = ½gt², so h ∝ t². That is why the h–t graph curves upwards as a parabola.

h ∝ t²h₁ / h₂ = t₁² / t₂²

The free fall formulas and where they come from

Take the equations for constant acceleration, set a = g and s = h, and for a dropped object put u = 0. These four formulas solve every free fall problem.

v = u + gtdropped from rest: v = gt

h = ut + ½gt²dropped from rest: h = ½gt²

v² = u² + 2ghdropped from rest: v = √(2gh)

t = √(2h/g)free fall formula for time

Why the velocity is v = gt

Acceleration is the gain in velocity per second. Gaining g every second for t seconds adds g × t in total. The object started at zero, so after t seconds its velocity is gt.

Why the height is h = ½gt²

The velocity rose steadily from 0 to gt, so the average velocity over that time is halfway between: (0 + gt) ÷ 2 = ½gt. Distance = average velocity × time = ½gt × t = ½gt². It is also the area of the triangle under the v–t graph: ½ × base t × height gt.

The third equation and the time to fall

Eliminate t from the first two and you get v² = 2gh, which is the one to use when you know the height but not the time. Rearranging h = ½gt² for t gives t = √(2h/g). Here is the phone puzzle solved: t ∝ √h, so doubling the height multiplies the time by √2 ≈ 1.41, not by 2.

What g = 9.8 m/s² actually means

g = 9.8 m/s² means a freely falling object gains 9.8 m/s of velocity every second. The internationally agreed standard value is 9.80665 m/s², but school problems use 9.8 (and some use 10 for quick estimates; read the question).

g is the same for every object at one place, but it is not the same at every place. Earth is slightly flattened at the poles, so an object there is a little closer to Earth’s centre: g is about 9.83 m/s² at the poles and 9.78 m/s² at the equator. It also falls slightly as you climb a mountain or go deep into a mine.

On another world g changes a lot, because it depends on the planet’s mass and radius. The table works out the same 20 m drop as the simulation’s default on each world.

WorldgTime to fall 20 mSpeed on landing
Earth9.80 m/s²2.02 s19.8 m/s
Moon1.62 m/s²4.97 s8.0 m/s
Mars3.71 m/s²3.28 s12.2 m/s
Jupiter24.79 m/s²1.27 s31.5 m/s

Adding air: why a feather falls slowly

In real life a feather drifts and a sheet of paper flutters. The reason is air, not gravity. An object moving through air is pushed back by the air it has to shove aside. That drag grows as the speed grows, and it matters most for objects with a large surface and a small mass.

At the moment of release the drag is almost zero, so even a feather starts falling at nearly g. As it speeds up the drag grows, until drag equals weight. Then the net force is zero, the acceleration is zero, and the speed stops increasing. That steady speed is the terminal velocity. The simulation’s feather settles at about 1.5 m/s and its coin near 45 m/s, which is why the coin hardly notices the air over a short drop.

Take the air away and they fall together, and this has been shown for real. In the eighteenth century the “guinea and feather” demonstration dropped a gold coin and a feather together in a glass tube pumped free of air. In 1971 the Apollo 15 astronaut David Scott stood on the Moon, where there is no air, and dropped a hammer and a falcon feather at the same time. They landed together, exactly as Galileo predicted.

Energy in free fall: potential turns into kinetic

An object held up high stores gravitational potential energy, mgh. As it falls its height drops, so its potential energy drops. The energy is not lost: its kinetic energy ½mv² rises by exactly the same amount. Without air, potential energy + kinetic energy stays constant at every instant. That is conservation of energy.

It gives a second route to the landing speed: all of the starting mgh becomes ½mv² at the bottom, so mgh = ½mv² and v = √(2gh). The mass cancels on both sides, which shows once again that the landing speed does not depend on mass. With air, some energy goes into pushing and warming the air, so a feather lands much more slowly.

Experiments to try in the simulation

Grab a notebook. Predict each result first, then run it and check.

  • With the height at 20 m, read “Formula time”, then run the drop and read “Coin landed at”. Both should be 2.02 s.
  • Change the height from 20 m to 80 m (four times as high). How many times longer is the fall? Twice as long, because t ∝ √h.
  • With the strobe on, read the ratios beside the coin: 1, 3, 5, 7. Turn the air on and notice that the coin’s numbers barely change.
  • Turn the air on and watch the feather’s line on the v–t graph: steep at first, then flat. The flat part is the terminal velocity.
  • Pick Jupiter: it all happens so fast you will want 0.25× speed. Then try Mars with the air on; its air is so thin that even the feather almost keeps up with the coin.

Solved free fall problems

For each problem, write down what is given, then the formula, then substitute. That is also how to lay out an exam answer. Air resistance is ignored throughout.

Problem 1: dropped from 20 m (the simulation’s default)

Given: h = 20 m, u = 0, g = 9.8 m/s². Find the time to reach the ground and the landing velocity.

t = √(2h/g) = √(40 / 9.8) = √4.08 = 2.02 s. v = √(2gh) = 19.80 m/s. Open the simulation and the readings panel shows exactly these values.

Problem 2: velocity and distance after 3 s

A stone is dropped from a roof. What is its velocity after 3 s, and how far has it fallen?

v = gt = 9.8 × 3 = 29.4 m/s. h = ½gt² = ½ × 9.8 × 3² = 44.1 m.

Problem 3: how deep is the well?

A stone dropped into a well is heard hitting the water 2.5 s later (ignore the time the sound takes to travel back up). How deep is the water surface?

h = ½gt² = ½ × 9.8 × 2.5² = 30.625 m, and the stone hits the water at v = gt = 24.5 m/s. In reality the sound needs a moment to climb back, so the true depth is a little less.

Problem 4: a ball thrown straight up

A ball is thrown vertically upwards at 19.6 m/s. How long does it take to reach the top, how high does it go, and when does it return to the hand?

Going up, g slows it down. At the top v = 0, so 0 = u − gt and t = u/g = 19.6 / 9.8 = 2 s. Maximum height H = u²/2g = 19.6 m. The way down takes the same time, so the total is 4 s. The downward half is simply free fall.

Problem 5: the same drop on the Moon

On the Moon g = 1.62 m/s². How long does a 20 m drop take, and how fast does the object land?

t = √(2h/g) = 4.97 s and v = √(2gh) = 8.05 m/s. The fall takes √(9.8/1.62) = 2.46 times as long as on Earth. Pick “Moon” in the simulation and compare.

Problem 6: distance fallen in the third second

The distance fallen during the nth second is hₙ = ½g(2n − 1). With n = 3: h = 4.9 × 5 = 24.5 m.

Compare the first three seconds: 4.9 m, then 14.7 m, then 24.5 m, in the ratio 1 : 3 : 5, exactly like the strobe.

Problem 7: comparing velocities with the second law

A freely falling object has a velocity of 19.6 m/s after 2 s. What is its velocity after 6 s?

By the second law v₂/v₁ = t₂/t₁ = 6/2 = 3, so v₂ = 3 × 19.6 = 58.8 m/s. The ratio method works even when g is not given.

Problem 8: comparing distances with the third law

An object falls 19.6 m in the first 2 s. How far does it fall in the first 4 s?

By the third law h₂/h₁ = (t₂/t₁)² = (4/2)² = 4, so h₂ = 4 × 19.6 = 78.4 m. Twice the time, four times the distance.

Problem 9: landing speed from energy

A 2 kg object is dropped from 20 m. What is its potential energy at the start, and its speed on landing?

Potential energy = mgh = 2 × 9.8 × 20 = 392 J. At the ground all of it is kinetic: ½mv² = 392 J, so v = 19.80 m/s, the same answer as Problem 1.

At a glance: velocity and distance, second by second

An object falling from rest with g = 9.8 m/s². Look at the odd numbers in the last column.

TimeVelocity, v = gtTotal fallen, h = ½gt²Fallen in that secondRatio
1 s9.8 m/s4.9 m4.9 m1
2 s19.6 m/s19.6 m14.7 m3
3 s29.4 m/s44.1 m24.5 m5
4 s39.2 m/s78.4 m34.3 m7
5 s49.0 m/s122.5 m44.1 m9

Common mistakes

Avoid these and free fall questions become easy marks.

  • Writing “heavier objects fall faster”. Without air resistance all objects fall together; any difference comes from air.
  • Dropping the conditions “from rest” and “without resistance” when stating the laws.
  • Assuming that doubling the height doubles the time. Since t ∝ √h, the time grows by √2.
  • Confusing the distance fallen in the nth second, ½g(2n − 1), with the total distance in the first n seconds, ½gn².
  • Getting the sign wrong for an object thrown upwards. If up is positive, g must be entered as negative.
  • Thinking the acceleration is zero at the highest point. The velocity is zero there; the acceleration is still g, downwards.
  • Writing the unit of g as m/s. It is an acceleration, so the unit is m/s².

Free fall in real life

The laws of free fall explain far more than exam questions.

  • Parachutes: a huge canopy multiplies the drag and lowers the terminal velocity enough for a safe landing.
  • Raindrops: they fall from kilometres up but reach a modest terminal velocity, so they do not hit like bullets.
  • Hard hats on building sites: even a small bolt dropped from high up lands fast, because v = √(2gh).
  • Estimating depth: drop a stone into a well or gorge, time it, and use h = ½gt².
  • Catching a ball: a fielder judging where a high catch will come down is solving free fall in their head.
  • Weightlessness in orbit: a space station and its crew are falling towards Earth together, which is why the astronauts float. That is free fall too.

Exam corner

Free fall appears in Grade 9 gravitation and Grade 11 kinematics, in SSC and HSC physics, and in GCSE and IGCSE courses. Expect four kinds of question: state the laws of falling bodies; explain what g = 9.8 m/s² means or why g varies from place to place; calculate time, height or velocity; and interpret v–t and h–t graphs.

A typical structured question

Riya drops a marble and a scrap of paper together from a 20 m roof. The marble lands first.

(a) Define acceleration due to gravity. (b) State the first law of falling bodies. (c) Calculate the time the marble takes to land. (d) “The paper landing later proves the first law wrong.” Evaluate this statement.

Answer to (c): t = √(2h/g) = 2.02 s. The key to (d): the law assumes no resistance. The paper has a large area and little mass, so air resistance slows it far more. Without air, as on the Moon, the two would land together, so the statement is wrong.

Revision summary

Read this list and the table above the night before the exam.

  • Free fall: motion under gravity alone, from rest, with no resistance.
  • g ≈ 9.8 m/s², the same for every object at one place; larger at the poles, smaller at the equator.
  • First law: equal distances in equal times. Second law: v ∝ t. Third law: h ∝ t².
  • Formulas: v = gt, h = ½gt², v² = 2gh, t = √(2h/g).
  • Distances in equal time steps go 1 : 3 : 5 : 7; in the nth second the fall is ½g(2n − 1).
  • With air, light and broad objects reach a terminal velocity; without air a feather and a coin fall together.
  • Falling converts potential energy into an equal amount of kinetic energy.

Frequently asked questions

What is free fall?

Free fall is motion under the influence of gravity alone, with no air resistance or other force. Near Earth every freely falling object accelerates downwards at about 9.8 m/s², whatever its mass.

What is the free fall formula for time?

For an object dropped from rest, t = √(2h/g), where h is the height and g the acceleration due to gravity. It comes from h = ½gt².

What is the free fall formula for velocity?

v = gt if you know the time, or v = √(2gh) if you know the height. Both assume the object starts from rest and air resistance is ignored.

Do heavier objects fall faster?

Not without air. Gravity pulls a heavier object harder, but its greater inertia needs exactly that much more force, so the acceleration is the same. In air, light objects with a large surface are slowed more by drag.

What are the three laws of falling bodies?

For bodies falling from rest without resistance: all bodies cover equal distances in equal times; the velocity is proportional to time (v ∝ t); and the distance is proportional to the square of time (h ∝ t²).

What does g = 9.8 m/s² mean?

It means a freely falling object gains 9.8 m/s of velocity every second: 9.8 m/s after 1 s, twice that after 2 s, and so on.

What is terminal velocity?

It is the constant speed a falling object reaches when air resistance has grown to equal its weight. The net force and acceleration are then zero, so the speed stops rising.

Is an astronaut in orbit in free fall?

Yes. The station and the astronauts are all falling towards Earth under gravity alone, while moving sideways fast enough to keep missing it. Falling together, nothing presses on anything, which is why they feel weightless.

What is the acceleration at the highest point of an upward throw?

The velocity is momentarily zero there, but the acceleration is still g = 9.8 m/s² downwards. If it were zero, the object would stay at the top instead of falling back.

Keep studying this topic

The animation made the idea click; now turn it into marks. Syllabus, suggestions, textbooks and admission-test guides are below.

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