Free-Fall Calculator

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

Dropped from rest	$t=\sqrt{\frac{2h}{g}},\qquad v=\sqrt{2gh}$
Thrown with v₀, up positive	$y=h+v_0t-\tfrac12gt^{2},\qquad v_y=v_0-gt$
Energy per kilogram, any throw	$\tfrac12v^{2}=\tfrac12v_0^{2}+gh$

Worked example — Free-Fall Calculator

Fall time	4.5152 s
Impact speed	44.29 m/s
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 in a vacuum every figure the panel reports follows from that single assumption. The air setting adds back the one force it leaves 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 44.29 m/s.
v₀: the launch velocity, positive upwards. A stone dropped has v₀ = 0; one thrown down has a negative v₀.
v: the impact speed at the moment it lands, not an average. A dropped 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

Measure the height up from the ground, with up positive. Gravity is the only force, so the acceleration is −g for the whole flight and the height follows the constant-acceleration equation.
$a_y=-g\ \Rightarrow\ v_y=v_0-gt,\quad y=h+v_0t-\tfrac12gt^{2}$
Dropped from rest, v₀ = 0, and the ground is y = 0. Solve for the time. It grows as the square root of the height, which is why a fall from twice the height does not take twice as long.
$0=h-\tfrac12gt^{2}\ \Rightarrow\ t=\sqrt{\frac{2h}{g}}$
The speed at that instant is gt. Put in the time and the height and the gravity enter symmetrically, while the mass never appears.
$v=gt=g\sqrt{\frac{2h}{g}}=\sqrt{2gh}$
Square it and halve it: the kinetic energy per kilogram is exactly gh. That is the potential energy the drop gave up, with the mass divided out of both sides, and it is why 100 m at 9.81 m/s² reads 981 J/kg.
$\tfrac12mv^{2}=mgh\ \Rightarrow\ \frac{E_k}{m}=\tfrac12v^{2}=gh$
Thrown, the energy equation still holds, with the energy of the launch speed added. The direction of the throw is not in it, because v₀ is squared: up or down at the same speed, the stone lands at the same speed. The time does depend on the direction, and comes from v = v₀ − gt with the landing velocity pointing down.
$\tfrac12v^{2}=\tfrac12v_0^{2}+gh\ \Rightarrow\ v=\sqrt{v_0^{2}+2gh},\qquad -v=v_0-gt\ \Rightarrow\ t=\frac{v_0+v}{g}$

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 140.07 m/s against 44.29 m/s — both multiplied by 3.1623, because ten times the height is √10 times the time and the speed, while the energy per kilogram goes up the full ten times. Quarter the height instead and the speed halves exactly, because v goes as √h: 25 m gives 22.15 m/s.

Assumes: No air, and a gravity that does not change over the drop. The second is the milder: 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 breaks it first, and completely rather than slightly. A skydiver stops accelerating near 55 m/s, so the vacuum setting overstates the 1000 m drop by more than a factor of two at 140.07 m/s. Nothing in v = √(2gh) can produce a ceiling. Switch the air on and the same drop lands at 54.96 m/s, after 22.0658 s.

Problems 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 the halfway height nor the halfway energy 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.
  
  $y(t)=h-\tfrac{1}{2}gt^{2}=0\quad\Longrightarrow\quad t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{200}{9.81}}=4.5152\ \mathrm{s}$
2. Speed is the derivative, evaluated at that instant. It is worth converting: 159 km/h, faster than most motorway limits, from about thirty storeys up.
  
  $v=gt=9.81\times4.5152=44.29\ \mathrm{m/s}=159.5\ \mathrm{km/h}$
3. Substituting v² = 2gh into the kinetic energy per kilogram makes the speed disappear, leaving the drop height and gravity. This is potential energy handed straight back, which is why the number is a round 981.
  
  $v^{2}=2gh\quad\Longrightarrow\quad \frac{E_k}{m}=\tfrac{1}{2}v^{2}=gh=9.81\times100=981\ \mathrm{J/kg}$
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.
  
  $\frac{y}{h}=1-\left(\frac{t}{t_{\text{imp}}}\right)^{2}:\quad t=\tfrac{1}{2}t_{\text{imp}}\Rightarrow\text{only }25\%\text{ fallen};\quad y=\tfrac{1}{2}h\Rightarrow t=0.707\,t_{\text{imp}}$
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.
  
  $v_t=55\ \mathrm{m/s}:\quad v_{\text{imp}}=v_t\sqrt{1-e^{-2gh/v_t^{2}}}=55\sqrt{1-e^{-0.6486}}=38.0\ \mathrm{m/s},\quad \frac{E_k}{m}=721.8$
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, 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.

$t=4.5152\ \mathrm{s},\quad v=44.29\ \mathrm{m/s},\quad E_k/m=gh=981\ \mathrm{J/kg}$
How deep is the well? 3 steps
A stone dropped into a well hits the water 2.5 s later. Ignoring the time the sound takes to come back up, how far down is the water, and how fast is the stone moving when it hits?

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1. Up positive, the water at y = 0, and the stone released from rest at the unknown height h. This time the known quantity is the time:
  
  $y=h-\tfrac12gt^{2}=0,\qquad t=2{.}5\ \mathrm{s}$
2. Solve for the height:
  
  $h=\tfrac12gt^{2}=\tfrac12(9{.}81)(2{.}5)^{2}=30{.}656\ \mathrm{m}$
3. The speed follows from the time, and energy gives the same number without it:
  
  $v=gt=9{.}81\times2{.}5=24{.}525\ \mathrm{m/s},\qquad v=\sqrt{2gh}=\sqrt{2(9{.}81)(30{.}656)}=24{.}525\ \mathrm{m/s}\ \checkmark$
Answer

30.66 m down, hit at 24.53 m/s. The sound needs about 0.09 s to climb back at 343 m/s, so a stopwatch that reads 2.5 s from release to splash means a well slightly shallower than this.

$h=30{.}66\ \mathrm{m},\qquad v=24{.}53\ \mathrm{m/s}$
Thrown up from a cliff 4 steps
A stone is thrown straight up at 25 m/s from the edge of a cliff 50 m high, and on the way down it misses the edge and falls to the ground below. How high above the ground does it rise, when does it land, and how fast is it going?

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1. Up positive and the ground at y = 0, so h = 50 m and v₀ = +25 m/s. At the top the velocity is zero:
  
  $0=v_0-g\,t_{\mathrm{top}}\ \Rightarrow\ t_{\mathrm{top}}=\frac{25}{9{.}81}=2{.}5484\ \mathrm{s},\qquad h_{\max}=h+\frac{v_0^{2}}{2g}=50+\frac{25^{2}}{2(9{.}81)}=81{.}855\ \mathrm{m}$
2. The landing speed comes from energy, without the time:
  
  $v=\sqrt{v_0^{2}+2gh}=\sqrt{25^{2}+2(9{.}81)(50)}=\sqrt{1606}=40{.}075\ \mathrm{m/s}$
3. The landing velocity points down, so it is −40.075 m/s, and v = v₀ − gt gives the time:
  
  $-40{.}075=25-9{.}81\,t\ \Rightarrow\ t=\frac{25+40{.}075}{9{.}81}=6{.}6335\ \mathrm{s}$
4. Throw it down at the same speed instead: v₀ = −25 m/s. The energy line does not change, because v₀ is squared. The time does:
  
  $v_0=-25:\quad v=\sqrt{(-25)^{2}+2(9{.}81)(50)}=40{.}075\ \mathrm{m/s},\qquad t=\frac{-25+40{.}075}{9{.}81}=1{.}5367\ \mathrm{s}$
Answer

81.86 m above the ground after 2.5484 s; it lands after 6.6335 s at 40.07 m/s. Thrown down at 25 m/s it lands after only 1.5367 s, at the same 40.07 m/s. The direction of the throw changed the time and the highest point, never the landing speed. Press Thrown up from 50 m and play it.

$h_{\max}=81{.}86\ \mathrm{m}\ (t=2{.}5484\ \mathrm{s}),\qquad t=6{.}6335\ \mathrm{s},\qquad v=40{.}07\ \mathrm{m/s}$
Finding g on the Moon 2 steps
An astronaut on the Moon drops a hammer from 1.6 m, and it reaches the ground 1.41 s later. What is the Moon's gravitational acceleration, and how fast does the hammer land?

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1. Dropped from rest, so the height equation has one unknown, g:
  
  $h=\tfrac12gt^{2}\ \Rightarrow\ g=\frac{2h}{t^{2}}=\frac{2(1{.}6)}{1{.}41^{2}}=1{.}6096\ \mathrm{m/s^{2}}$
2. The landing speed from the time, checked by energy:
  
  $v=gt=1{.}6096\times1{.}41=2{.}2695\ \mathrm{m/s},\qquad \sqrt{2gh}=\sqrt{2(1{.}6096)(1{.}6)}=2{.}2695\ \mathrm{m/s}\ \checkmark$
Answer

g = 1.61 m/s², and the hammer lands at 2.27 m/s. On Earth the same drop takes 0.5711 s and lands at 5.6 m/s. A feather dropped beside the hammer on the Moon lands at the same moment, because nothing in either line contains the mass.

$g=1{.}61\ \mathrm{m/s^{2}},\qquad v=2{.}27\ \mathrm{m/s}$
The second half in the last second 4 steps
A stone dropped from rest covers the second half of its height in the last 1.0 s of its fall. How long does the whole fall take, and from what height was it dropped?

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1. Call the whole fall time T. At T − 1 the stone has fallen half the height, and the full height is ½gT²:
  
  $\tfrac12g(T-1)^{2}=\tfrac12\cdot\tfrac12gT^{2}$
2. g cancels, so the answer for the time holds on any planet. Take the square root:
  
  $(T-1)^{2}=\frac{T^{2}}{2}\ \Rightarrow\ T-1=\pm\frac{T}{\sqrt2}$
3. The minus sign gives T = 0.5858 s, shorter than the 1 s the question is about, so that root is thrown away:
  
  $T=\frac{1}{1-1/\sqrt2}=3{.}4142\ \mathrm{s}\qquad\left(T=\frac{1}{1+1/\sqrt2}=0{.}5858\ \mathrm{s}<1\ \mathrm{s}\ \times\right)$
4. Then the height:
  
  $h=\tfrac12gT^{2}=\tfrac12(9{.}81)(3{.}4142)^{2}=57{.}177\ \mathrm{m}$
Answer

3.4142 s, from 57.18 m. The time does not depend on g, so the same condition on the Moon also means a 3.4142 s fall, from only 9.44 m.

$T=3{.}4142\ \mathrm{s},\qquad h=57{.}18\ \mathrm{m}$
Where the energy goes 3 steps
A skydiver falls 1000 m from rest through air that limits her to a terminal speed of 55 m/s. How fast is she falling at the end of the 1000 m, and what share of the energy gravity supplied per kilogram has gone into the air?

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1. Without air all of it would become speed: gh per kilogram.
  
  $gh=9{.}81\times1000=9810\ \mathrm{J/kg},\qquad \sqrt{2gh}=140{.}07\ \mathrm{m/s}$
2. With drag proportional to v², the speed after a drop h from rest has a closed form:
  
  $v=v_t\sqrt{1-e^{-2gh/v_t^{2}}}=55\sqrt{1-e^{-6{.}4860}}=54{.}958\ \mathrm{m/s}$
3. Her kinetic energy per kilogram, and the remainder, which the drag took as negative work:
  
  $\tfrac12v^{2}=\tfrac12(54{.}958)^{2}=1510{.}19\ \mathrm{J/kg},\qquad \frac{W_{\mathrm{air}}}{m}=9810-1510{.}19=8299{.}81\ \mathrm{J/kg}$
Answer

54.96 m/s, and 85% of the energy, 8,299.81 of 9,810 J/kg, has gone into the air. She was within 1% of the terminal speed after about 604 m, and nearly all of every metre after that went into the air instead of into her speed. Press Skydiver in air, 1000 m to see the speed curve flatten.

$v=54{.}96\ \mathrm{m/s},\qquad \frac{W_{\mathrm{air}}}{gh}=\frac{8299{.}81}{9810}=85\%$

Supplementary problems, answers only

A ball is dropped from a window 20 m above the ground. How long does it fall, and how fast does it land?

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2.0193 s and 19.81 m/s.

$t=\sqrt{\frac{2(20)}{9{.}81}}=2{.}0193\ \mathrm{s},\qquad v=\sqrt{2(9{.}81)(20)}=19{.}81\ \mathrm{m/s}$
A ball is thrown straight up from the ground at 15 m/s. How high does it rise, and when does it come back down?

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11.47 m, and it is back after 3.0581 s.

$h_{\max}=\frac{15^{2}}{2(9{.}81)}=11{.}47\ \mathrm{m},\qquad t=\frac{2(15)}{9{.}81}=3{.}0581\ \mathrm{s}$
A stone is thrown straight down at 10 m/s from a bridge 40 m above the water. How fast does it hit the water, and when?

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29.75 m/s, after 2.0128 s.

$v=\sqrt{10^{2}+2(9{.}81)(40)}=29{.}75\ \mathrm{m/s},\qquad t=\frac{-10+29{.}746}{9{.}81}=2{.}0128\ \mathrm{s}$
How long does a 10 m drop take on Mars, where g = 3.72 m/s², and how fast does it land?

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2.3187 s and 8.63 m/s.

$t=\sqrt{\frac{2(10)}{3{.}72}}=2{.}3187\ \mathrm{s},\qquad v=\sqrt{2(3{.}72)(10)}=8{.}63\ \mathrm{m/s}$
A stone dropped from rest falls 25 m in the last second before it lands. From what height was it dropped?

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45.58 m: the fall lasts 3.0484 s.

$\tfrac12gT^{2}-\tfrac12g(T-1)^{2}=25\ \Rightarrow\ T=\frac{25}{9{.}81}+\tfrac12=3{.}0484\ \mathrm{s},\qquad h=\tfrac12gT^{2}=45{.}58\ \mathrm{m}$

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, 44.29 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 44.29 m/s, against Earth's 981 and the same 44.29. 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?

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Between 154 m and 155 m: 154 gives 54.97 m/s and 155 gives 55.15 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 140.07 m/s, two and a half times a speed a falling body in air cannot pass at all.

Learning path

Energy instead of forces

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 7 Raven Room team · Reviewed against published sources. Corrections welcome.