Physics · Kinematics

Free fall: vertical motion under gravity

Q: How do you calculate the fall time from the height?

From rest at constant g, $t = \sqrt{(\frac{2h}{g})}$. A 12 m drop takes about 1.56 s with $g =\frac{ 9.81 m}{s^{2}}$. Four times the height takes twice the time. Timing until hearing impact also includes sound travel $\frac{h}{c}$; that measured interval must not automatically be treated as pure fall time. Choose the positive root because elapsed time is required. Then substitute it into $\frac{1}{2}gt^{2} = h$ as a check. The quantity inside the square root has units $s^{2}$. For zero initial velocity this calculation is unambiguous; a vertical throw instead requires solving the full motion equation with its linear term retained.

Q: At what speed does a falling body hit the ground?

For release from rest without drag, speed is $v = gt = \sqrt{(2gh)}$. At 12 m with $g =\frac{ 9.81 m}{s^{2}}$ it is $\frac{15.34 m}{s}$. Motion is downward; with upward positive the component is therefore $-\frac{15.34 m}{s}$. Here $\frac{h}{t}$ is only half the final speed. Nonzero v₀ or drag changes the equations. Check the energy calculation: lost gravitational potential energy mgh equals $\frac{1}{2}mv^{2}$ in this model. Mass cancels under the stated assumptions. Velocity immediately before contact also does not determine impact force. That force depends, among other factors, on the duration and distance over which contact stops the body, which the falling equation does not specify.

Q: What changes for a fall with initial velocity or on other planets?

With downward positive, $h = v_{0}t + \frac{1}{2}gt^{2}$. For an upward throw v₀ is then negative; alternatively choose upward positive and $a = -g$. At the peak $v = 0$ but a remains nonzero. On other celestial bodies use the appropriate local g. A constant-g model remains suitable only when field variations are sufficiently small. For an original launch at $+\frac{14 m}{s}$ with upward positive, the peak occurs after $\frac{14}{9.81} \approx{} 1.43 s$. Velocity then becomes negative and the body falls back. Returning to the same launch level takes twice the ascent time in the ideal model; this symmetry does not apply to a different final level.

Free fall from rest without air resistance: downward distance grows quadratically with time at constant g. Free fall with initial velocity also needs the v₀t term.

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Formula

Calculate

h = \frac{1}{2} g t^{2}

h in metres [m] · local g in m/s², often 9.81 m/s² near Earth · t in seconds [s]

Variables & units

Variables & units
Symbol	Meaning	Unit
$h$	Distance fallen since release from rest	m
$g$	Local magnitude of gravitational acceleration	m/s²
$t$	Time since release	s

LaTeX

h = \frac{1}{2} g t^2

QUANTA STUDYquanta-study.de

Original parabola h=½gt² for a fall from rest at constant g without air resistance. Downward displacement h increases with time. — For a fall from rest without air resistance and with constant g, the fallen height h grows quadratically with t. h is downward displacement, not height above the ground.

Free fall calculator

Find fall distance, final speed or fall time for a drop from rest, without air resistance.

Find

Gravitational acceleration g

Gravitational acceleration, 9.81 m/s² on Earth, about 1.62 m/s² on the Moon.

m/s²

Time t

Fall time from release at rest, e.g. 2 s.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for h, substitute, evaluate

1. Rearrange for h
$h=\frac12gt^2$
2. Substitute values (in SI units)
$h=\frac12\cdot 9.81\cdot 2^2=19.62$
3. Result
h = 19.62 m

Result: Downward fall distance h

h = 19.62 m

With g = 9.81 m/s² and t = 2 s, h = 19.62 m.

Quick tips

Estimate with g ≈ 10 m/s²: after 1 s about 5 m fallen, after 2 s about 20 m, after 3 s about 45 m.
Twice the fall time means four times the distance: h grows with t², the speed only with t.

Assumptions and limits

Zero initial velocity, constant g > 0 and no air drag. h is downward distance and v is speed magnitude; only t ≥ 0.

Derivation & background

Bodies with equal initial conditions in the same field fall independently of mass when air resistance is absent. The Apollo 15 hammer-and-feather demonstration illustrates this in the nearly airless lunar environment. The displayed short form requires $v_{0} = 0$ and constant g. Impact speed is then $gt = \sqrt{(2gh)}$ .

Exam blueprint

Validity range

$h = \frac{1}{2}gt^{2}$ describes a drop from rest at approximately constant g without drag or other forces. General free fall can have $v_{0} \ne{} 0$ and is not limited to vertical release. $g =\frac{ 9.81 m}{s^{2}}$ is a rounded near-Earth model value.

Derivation steps

Insert a drop from rest at constant g as a specific kinematic case.

1Choose downward as positive and the release point as initial position.
2Without drag or additional forces, acceleration component $a = g$ .
3Assume constant g and release from rest: $v_{0} = 0$ .
4Substitute into s = v₀t + ½at²: $h = \frac{1}{2}gt^{2}$ . h is distance fallen since release.
5The velocity equation gives vdown = gt and hence vdown² = 2gh.
6For time $t = \sqrt{(\frac{2h}{g})}$ , choose the nonnegative root and check units and the specified g value.

Rearrangements

Fall time from the height

t = \sqrt{\frac{2h}{g}}

The fall time grows only with the square root of the height.

Impact speed from the height

v = \sqrt{2gh}

Also follows directly from energy conservation mgh = ½mv².

Task variant

A stone is released from rest at a 45 m cliff. Neglect air forces and use $g =\frac{ 9.81 m}{s^{2}}$ . Find time until just before impact.

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{90}{9.81}}\,\mathrm s\approx\sqrt{9.17}\,\mathrm s\approx3.0\,\mathrm s$ .

Find the speed 1.5 s after release from rest. Gravity is the only force and $g =\frac{ 9.81 m}{s^{2}}$ remains constant.

v = g·t = 9.81 × 1.5 ≈ 14.7 m/s (about 53 km/h).

Common mistakes

Assuming heavy bodies fall faster.

With identical initial conditions in the same field, free-fall acceleration is mass-independent when gravity is the only force. Compare equal times after release to obtain equal velocities.

Equating double fall time with double height.

Only for release from rest at constant g is distance fallen proportional to $t^{2}$ . Under those assumptions, twice the time means four times the falling distance.

Linking fall time and impact speed via $v =\frac{ h}{t}$ .

$\frac{h}{t}$ is only the average speed; at the ground $v = g\cdot{}t$ holds.

Exam context

Typical entry into energy-conservation and projectile tasks; often combined with reaction time or the speed of sound (well problems).

Formula cluster

Falling and projectile motion

Connects kinematics with energy conservation in the gravitational field.

s = ½at² + v₀t

Uniformly Accelerated Motion (Position–Time Equation)

v = a·t + v₀

Velocity-Time Law

Epot = mgh

Potential Energy (Gravitational)

x = v₀·cosα·t; y = v₀·sinα·t − ½gt²

Projectile Motion (Trajectory Parabola)

Worked example

A stone is released from rest from a 19.62 m bridge. With no air forces and constant $g=9.81\,\mathrm{m/s^2}$ , it reaches the ground after 2 s. Approach: $h=\frac12gt^2$ , $v=gt$ . Substitution gives $h=\frac12(9.81)(2)^2\,\mathrm m=19.62\,\mathrm m$ and, just before contact, $v=9.81(2)\,\mathrm{m/s}=19.62\,\mathrm{m/s}\approx70.6\,\mathrm{km/h}$ .

Applications

Drop towers for microgravity, well depth by stopwatch, stunt planning, material drop tests

How do you calculate with Free Fall?

Here is how to work through a typical Free Fall (h = ½gt²) task step by step:

1
Task
A stone is released from rest at a 45 m cliff. Neglect air forces and use $g =\frac{ 9.81 m}{s^{2}}$ . Find time until just before impact.
Solution path
$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{90}{9.81}}\,\mathrm s\approx\sqrt{9.17}\,\mathrm s\approx3.0\,\mathrm s$ .
2
Task
Find the speed 1.5 s after release from rest. Gravity is the only force and $g =\frac{ 9.81 m}{s^{2}}$ remains constant.
Solution path
v = g·t = 9.81 × 1.5 ≈ 14.7 m/s (about 53 km/h).

Understand and apply the equation

Why do successive positions of a falling object become farther apart? In ideal free fall, acceleration stays constant while velocity changes. We distinguish distance fallen from height above the ground, derive the equations step by step and also examine an upward launch.

Getting an intuitive picture

When you release a small compact object, it moves only a short distance during the first brief interval. Later it moves faster and covers larger distances in equal intervals. Our initial model neglects air forces and assumes constant gravitational acceleration. “Free” means that gravity is the only force after release. It does not mean the object is already moving fast or can only travel downward.[1]

Prerequisite: the constant-acceleration velocity and position equations apply. Starting from rest means v₀ = 0 in the selected reference frame.
g is a positive magnitude. The acceleration component is +g on a downward-positive axis and −g on an upward-positive axis.
The numerical examples use g = 9.81 m/s². This is a rounded model value; a problem may specify a different value.

Which free-fall equations apply to a drop from rest and a vertical launch?

In ideal vertical free fall with constant g, gravity is the only force during flight; a nonzero initial velocity is allowed. For a drop from rest, h denotes distance fallen since release. Height above the ground instead equals initial height minus h. A problem may use “height” for either quantity; establish which one is meant before substituting. More generally, the freely moving part of a vertical throw is also free fall when only gravity acts. Its initial velocity must not be omitted.[1]

\begin{aligned}h&=\frac12gt^2\\v_{\mathrm{down}}&=gt\quad(v_0=0)\end{aligned}

This notation takes the positive axis downward.

These problems use g = 9.81 m/s², a rounded near-surface model value rather than a universally exact constant. Measurement analysis needs local g; large height changes need a varying gravitational-field model.[1]

View diagram: direction of gravity

The force remains downward

QUANTA STUDYquanta-study.de

After release, ideal free fall involves weight alone. The arrow represents force, not velocity: it points downward even while a thrown object moves upward.[1]

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Fall velocity and fall time from the motion model

For the first case choose a downward-positive axis. Starting from rest, velocity grows as v = gt. The velocity graph is a line through the origin. Distance fallen is the triangle beneath it: half its base t times its height gt. This explains the factor one half. A nonzero initial velocity adds the rectangular contribution v₀t. This falling law is a special case of motion with constant acceleration, with downward-positive $a=g$ .[2]

\begin{aligned}h(t)&=\frac12t\,v(t)=\frac12t(gt)\\&=\frac12gt^2\\h(t)\big|_{v_0\ne0}&=v_0t+\frac12gt^2\end{aligned}

The first line assumes release from rest. In the last line h is signed displacement; it can initially be negative for an upward launch.[2][1]

\begin{aligned}t&=\sqrt{\frac{2h}{g}}\\v_{\mathrm{impact}}&=\sqrt{2gh}\quad(h\ge0,\ g>0)\end{aligned}

For release from rest choose the nonnegative time root; vimpact is a speed.

Quadrupling the drop distance doubles fall time and impact speed. Doubling fall time quadruples distance. From rest, average speed h/t is half the final speed and cannot be used as the impact value.

Distance fallen and height above ground

QUANTA STUDYquanta-study.de

For the 12 m drop from rest, distance fallen grows as remaining height decreases. Their sum is always 12 m. Both curves end at first ground contact.[1][2]

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Fall time and impact speed: a twelve-metre example

Time and speed from drop height

Given: Drop distance 12 m, release from rest, g = 9.81 m/s², negligible air resistance.
Find: Find time to the ground and speed immediately before impact.

Approach: distance fallen is known but time is not. Rearrange the drop equation for positive elapsed time, then obtain speed from gt.
$\begin{aligned}h&=\frac12gt^2\\t&=\sqrt{\frac{2h}{g}}\\v_{\mathrm{down}}&=gt\end{aligned}$
Substitute height into the time equation.
$t=\sqrt{\frac{2\cdot12}{9.81}}=1.56412\ldots\,\mathrm s$
Use the unrounded time to calculate speed.
$v=9.81\cdot1.56412\ldots=15.34405\ldots\,\mathrm{m/s}$
Optionally convert to km/h after the SI calculation.
$v\approx15.34405\cdot3.6=55.23860\,\mathrm{km/h}$

Result: Fall time 1.56 s; impact speed 15.34 m/s or 55.24 km/h.

Check: Independent energy check: $v^2=2gh=235.44\,\mathrm{m^2\,s^{-2}}$ . Also, $\frac{12\,\mathrm m}{1.56412\,\mathrm s}=7.67203\,\mathrm{m\,s^{-1}}$ , half the final speed.

Upward launch: is acceleration zero at the highest point?

Nonzero initial velocity

Given: Start at z₀ = 0 m with v₀ = +14 m/s, positive axis upward; g = 9.81 m/s².
Find: Find time to the peak, height gain and return time to the launch level.

Approach: now choose upward positive. Acceleration is −g. The peak occurs when vertical velocity is zero; return follows from equal initial and final height.
$\begin{aligned}v_z(t)&=v_0-gt\\z(t)-z_0&=v_0t-\frac12gt^2\\v_z(t_{\mathrm{peak}})&=0\\z(t_{\mathrm{return}})&=z_0\end{aligned}$
With upward positive, acceleration is −g.
$\begin{aligned}v(t)&=14-9.81t\\{} z(t)&=14t-\tfrac12\cdot9.81t^2\end{aligned}$
At the peak v = 0, not a = 0.
$t_{\mathrm{peak}}=\frac{14}{9.81}=1.42712\ldots\,\mathrm s$
Calculate peak height without premature rounding.
$\Delta z_{\max}=\frac{14^2}{2\cdot9.81}=9.98981\ldots\,\mathrm m$
z = 0 has the launch root and the later return root.
$\begin{aligned}t(14-4.905t)&=0\\{} t_{\mathrm{return}}&=\frac{28}{9.81}=2.85423\ldots\,\mathrm s\end{aligned}$

Result: Peak at 1.43 s with a 9.99 m height gain; return to launch level at 2.85 s with v = −14 m/s.

Check: Ascent and descent take equal time in this model because initial and final levels coincide. Acceleration remains −9.81 m/s² at the peak.

Do heavier objects fall faster? An energy check in the ideal model

mgh=\frac12mv^2\quad\Longrightarrow\quad v^2=2gh\quad(m>0)

In the ideal model, lost gravitational potential energy becomes kinetic energy.[3]

The energy check relates a decrease in gravitational potential energy to an increase in kinetic energy. Mass cancels when both energy terms scale with the same mass. Bodies with matching initial conditions in the same field reach the same velocity. With air resistance, shape, area, air density and mass also affect motion, so equal final speed in air does not follow.[3][4]

The Apollo 15 hammer-and-feather demonstration on the Moon illustrates falling without appreciable air resistance. It does not justify neglecting drag for a sheet of paper in a terrestrial stairwell.[5]

Interpreting a drop experiment or well measurement

For a drop experiment, h plotted against t² is linear with slope g/2 if release from rest and origins are correct. Hand-timing reaction delay shifts recorded time; an imperfect release can alter v₀. Multiple heights help reveal model deviations.

Timing from release until hearing an impact in a well includes sound travel back upward. With an assumed constant sound speed c, T = √(2h/g) + h/c. Treating all of T as falling time overestimates depth. Drag and measurement conditions remain additional error sources.

T=t_{\mathrm{fall}}+t_{\mathrm{sound}}=\sqrt{\frac{2h}{g}}+\frac hc

h/c is in seconds; c must be specified as a separate model assumption.

Drag, terminal speed and varying gravity

Air drag opposes motion relative to the air. A body falling through still air can approach terminal speed when weight and opposing forces balance, stopping further speed increase. Acceleration is then not constantly g, and h = ½gt² overestimates distance after a given time for the same initial conditions.[4]

The simple model also neglects buoyancy and air motion. Gravity cannot be kept unchanged over large height changes. An orbiting satellite is also in free fall, but its tangential velocity means its trajectory is not described by a purely vertical distance ½gt².[1]

All heights and times here are model calculations. An impact calculation ends immediately before contact; the contact force is not determined by the falling equation alone. Additional information such as deformation distance or stopping duration is needed. The change in momentum and stopping duration determine the mean net force; a definite peak force additionally requires the force history.[6]

Five solved falling exercises

Calculate · Foundation

An object falls from rest for 1.2 s. Find distance and speed using g = 9.81 m/s².

Show hint

Square t for distance, not for speed.

Learning goal: Find fall distance and velocity from rest.

Show solution and calculation

Approach: release from rest gives h = ½gt² and downward speed v = gt.
At 1.2 s the distance fallen is 7.0632 m and speed is 11.772 m/s.

Check: average speed over the drop is half the final speed; multiplying it by 1.2 s gives the same distance. Earlier contact with the ground would terminate the model.

\begin{aligned}h&=\frac12(9.81\,\mathrm{m/s^2})(1.2\,\mathrm s)^2=7.0632\,\mathrm m\\v&=(9.81\,\mathrm{m/s^2})(1.2\,\mathrm s)=11.772\,\mathrm{m/s}\end{aligned}

[1][2]

Rearrange · Foundation

How long does a drop from rest through 5 m take without air resistance?

Show hint

Use the positive time root.

Learning goal: Infer fall time from drop height.

Show solution and calculation

Approach: solve h = ½gt² for t.
The positive root represents time since release.
A five-metre drop takes about 1.01 s.

Check: substitution into ½gt² returns 5 m. The quantity beneath the root has units of seconds squared, so taking its root yields seconds.

\begin{aligned}t&=\sqrt{\frac{2h}{g}}\\&=\sqrt{\frac{2(5\,\mathrm m)}{9.81\,\mathrm{m/s^2}}}\\&\approx1.010\,\mathrm s\end{aligned}

[1][2]

Transfer · Application

An object is thrown downward at 3 m/s. How far does it fall in 0.8 s?

Show hint

With downward positive, v₀ and a are both positive.

Learning goal: Include nonzero initial throwing velocity.

Show solution and calculation

Approach: the initial downward velocity contributes v₀t.
Add it to the acceleration contribution ½gt².
At 0.8 s these contributions are 2.4 m and 3.1392 m, giving 5.5392 m.

Check: the result must exceed the distance of a drop from rest lasting the same time, since downward motion already exists at the start.

\begin{aligned}h&=v_0t+\frac12gt^2\\&=(3\,\mathrm{m/s})(0.8\,\mathrm s)+\frac12(9.81\,\mathrm{m/s^2})(0.8\,\mathrm s)^2\\&=2.4\,\mathrm m+3.1392\,\mathrm m=5.5392\,\mathrm m\end{aligned}

[1][2]

Interpret · Foundation

An object is thrown vertically upward at 9.81 m/s in vacuum, with constant g=9.81 m/s². Positive is upward. State time, velocity and acceleration at the apex. Is the ascending phase also free fall?

Show hint

Free fall describes the acting force, not instantaneous motion direction.

Learning goal: Distinguish zero speed at the apex from zero acceleration.

Show solution and calculation

Approach: $v=v_0-gt$ reaches zero after 1 s.
Acceleration remains −9.81 m/s²; gravity acts both before and after the apex.
Ascending motion is also free fall once the throwing hand loses contact.

Check: velocity is positive just before the apex and negative just after. Without continued acceleration, reversal could not occur.

\begin{aligned}t_{\mathrm{apex}}&=\frac{v_0}{g}=1\,\mathrm s\\v_{\mathrm{apex}}&=0\\a&=-9.81\,\mathrm{m/s^2}\end{aligned}

[1][2]

Error analysis · Application

A stone is dropped from rest into a vertical shaft. Time from release to hearing impact is 2.00 s. c=340 m/s, g=9.81 m/s²; neglect drag and reaction time. A solution gives h=19.62 m. Is that too large or too small? Find corrected depth.

Show hint

Fall time is less than 2 s because sound also needs time.

Learning goal: Model measured delay as fall time plus sound travel time.

Show solution and calculation

Approach: $2\,\mathrm s=t_f+\frac hc$ with $h=\frac12gt_f^2$ .
The positive solution gives tf≈1.94540 s and h≈18.5634 m.
The original 19.62 m is too large because the entire delay was counted as fall time.

Check: sound travel of about 0.05460 s completes 2.00 s. This assumes sound traverses the same vertical distance at the supplied constant c.

\begin{aligned}T&=t_f+\frac{gt_f^2}{2c}\\t_f&=\frac{2T}{1+\sqrt{1+\frac{2gT}{c}}}\\h&=\frac12gt_f^2\end{aligned}

[1][2]

What to take away

First identify whether height means distance fallen or height above the ground, and whether initial velocity is zero.
Weight and acceleration point downward even at the peak. Only velocity is momentarily zero there for a vertical throw.
The ideal equations describe motion before contact. Air forces and sound travel time require separate modelling when relevant.

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 2 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Distinguished free fall from a drop from rest directly; made fall time, impact speed, the highest point and mass accessible through appropriate headings. Linked the general displacement law, potential energy and momentum change during contact.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added prerequisites and an intuitive introduction; clarified the derivation, worked-example approaches and exercise solutions, and added a summary.

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Quanta. Free Fall. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/freier-fall

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Frequently asked questions about Free Fall

How do you calculate the fall height with

h = \frac{1}{2}gt^{2}

For release from rest at constant g, square t and multiply by $\frac{g}{2}$ . With 2 s and $g =\frac{ 9.81 m}{s^{2}}$ , distance fallen is 19.62 m. This is distance since release, not remaining height. A throw with initial velocity requires the additional v₀t term. Check that h is zero at $t = 0$ . Height remaining above the ground instead equals initial height minus distance fallen. Use the value of g specified in the problem; $\frac{9.81 m}{s^{2}}$ is a rounded near-Earth value. During ground contact this equation stops applying because additional contact forces act on the object.

Do heavy bodies really not fall faster than light ones?+

With equal initial conditions in the same field and no air resistance, acceleration is independent of mass. Hammer and feather on the Moon illustrate the model. In air, drag and buoyancy act, and shape, area and mass affect motion. Equal terminal speed in air does not follow from the ideal falling equation. Even in the ideal model, bodies can have different velocities if one begins with an initial velocity or is released earlier. Compare equal elapsed times from a common release and equal initial conditions. A feather is not automatically suitable for an Earth-based test of this approximation, because its large drag can dominate the motion.

How do you calculate the fall time from the height?+

From rest at constant g, $t = \sqrt{(\frac{2h}{g})}$ . A 12 m drop takes about 1.56 s with $g =\frac{ 9.81 m}{s^{2}}$ . Four times the height takes twice the time. Timing until hearing impact also includes sound travel $\frac{h}{c}$ ; that measured interval must not automatically be treated as pure fall time. Choose the positive root because elapsed time is required. Then substitute it into $\frac{1}{2}gt^{2} = h$ as a check. The quantity inside the square root has units $s^{2}$ . For zero initial velocity this calculation is unambiguous; a vertical throw instead requires solving the full motion equation with its linear term retained.

At what speed does a falling body hit the ground?+

For release from rest without drag, speed is $v = gt = \sqrt{(2gh)}$ . At 12 m with $g =\frac{ 9.81 m}{s^{2}}$ it is $\frac{15.34 m}{s}$ . Motion is downward; with upward positive the component is therefore $-\frac{15.34 m}{s}$ . Here $\frac{h}{t}$ is only half the final speed. Nonzero v₀ or drag changes the equations. Check the energy calculation: lost gravitational potential energy mgh equals $\frac{1}{2}mv^{2}$ in this model. Mass cancels under the stated assumptions. Velocity immediately before contact also does not determine impact force. That force depends, among other factors, on the duration and distance over which contact stops the body, which the falling equation does not specify.

What changes for a fall with initial velocity or on other planets?+

With downward positive, $h = v_{0}t + \frac{1}{2}gt^{2}$ . For an upward throw v₀ is then negative; alternatively choose upward positive and $a = -g$ . At the peak $v = 0$ but a remains nonzero. On other celestial bodies use the appropriate local g. A constant-g model remains suitable only when field variations are sufficiently small. For an original launch at $+\frac{14 m}{s}$ with upward positive, the peak occurs after $\frac{14}{9.81} \approx{} 1.43 s$ . Velocity then becomes negative and the body falls back. Returning to the same launch level takes twice the ascent time in the ideal model; this symmetry does not apply to a different final level.

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More formulas: Mechanics

\sum \vec F_{\mathrm{ext}} = \vec F_{\mathrm{res}} = m \vec a

Newton's Second Law

F_G = G \cdot \frac{m_1 \cdot m_2}{r^2}

Newton's Law of Universal Gravitation

E_{kin} = \frac{1}{2} m v^2

Kinetic Energy

F = k \cdot x

Hooke's Law (Spring Force)

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Free fall: vertical motion under gravity

Free fall from rest without air resistance: downward distance grows quadratically with time at constant g. Free fall with initial velocity also needs the v₀t term.

BasicExam-relevantLower secondaryUpper secondary

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Symbol

Meaning

Unit

h

Distance fallen since release from rest

g

Local magnitude of gravitational acceleration

m/s²

t

Time since release

Derivation & background

Exam blueprint

Validity range

Derivation steps

Insert a drop from rest at constant g as a specific kinematic case.

1Choose downward as positive and the release point as initial position.
2Without drag or additional forces, acceleration component $a = g$ .
3Assume constant g and release from rest: $v_{0} = 0$ .
4Substitute into s = v₀t + ½at²: $h = \frac{1}{2}gt^{2}$ . h is distance fallen since release.
5The velocity equation gives vdown = gt and hence vdown² = 2gh.
6For time $t = \sqrt{(\frac{2h}{g})}$ , choose the nonnegative root and check units and the specified g value.

How do you calculate with Free Fall?

Here is how to work through a typical Free Fall (h = ½gt²) task step by step:

Task

A stone is released from rest at a 45 m cliff. Neglect air forces and use $g =\frac{ 9.81 m}{s^{2}}$ . Find time until just before impact.

Solution path

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{90}{9.81}}\,\mathrm s\approx\sqrt{9.17}\,\mathrm s\approx3.0\,\mathrm s$ .

Task

Find the speed 1.5 s after release from rest. Gravity is the only force and $g =\frac{ 9.81 m}{s^{2}}$ remains constant.

Solution path

v = g·t = 9.81 × 1.5 ≈ 14.7 m/s (about 53 km/h).

Getting an intuitive picture

Prerequisite: the constant-acceleration velocity and position equations apply. Starting from rest means v₀ = 0 in the selected reference frame.
g is a positive magnitude. The acceleration component is +g on a downward-positive axis and −g on an upward-positive axis.
The numerical examples use g = 9.81 m/s². This is a rounded model value; a problem may specify a different value.

Which free-fall equations apply to a drop from rest and a vertical launch?

\begin{aligned}h&=\frac12gt^2\\v_{\mathrm{down}}&=gt\quad(v_0=0)\end{aligned}

This notation takes the positive axis downward.

View diagram: direction of gravity

The force remains downward

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After release, ideal free fall involves weight alone. The arrow represents force, not velocity: it points downward even while a thrown object moves upward.[1]

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Fall velocity and fall time from the motion model

\begin{aligned}h(t)&=\frac12t\,v(t)=\frac12t(gt)\\&=\frac12gt^2\\h(t)\big|_{v_0\ne0}&=v_0t+\frac12gt^2\end{aligned}

The first line assumes release from rest. In the last line h is signed displacement; it can initially be negative for an upward launch.[2][1]

\begin{aligned}t&=\sqrt{\frac{2h}{g}}\\v_{\mathrm{impact}}&=\sqrt{2gh}\quad(h\ge0,\ g>0)\end{aligned}

For release from rest choose the nonnegative time root; vimpact is a speed.

Quadrupling the drop distance doubles fall time and impact speed. Doubling fall time quadruples distance. From rest, average speed h/t is half the final speed and cannot be used as the impact value.

Distance fallen and height above ground

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For the 12 m drop from rest, distance fallen grows as remaining height decreases. Their sum is always 12 m. Both curves end at first ground contact.[1][2]

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Fall time and impact speed: a twelve-metre example

Time and speed from drop height

Given: Drop distance 12 m, release from rest, g = 9.81 m/s², negligible air resistance.
Find: Find time to the ground and speed immediately before impact.

Approach: distance fallen is known but time is not. Rearrange the drop equation for positive elapsed time, then obtain speed from gt.
$\begin{aligned}h&=\frac12gt^2\\t&=\sqrt{\frac{2h}{g}}\\v_{\mathrm{down}}&=gt\end{aligned}$
Substitute height into the time equation.
$t=\sqrt{\frac{2\cdot12}{9.81}}=1.56412\ldots\,\mathrm s$
Use the unrounded time to calculate speed.
$v=9.81\cdot1.56412\ldots=15.34405\ldots\,\mathrm{m/s}$
Optionally convert to km/h after the SI calculation.
$v\approx15.34405\cdot3.6=55.23860\,\mathrm{km/h}$

Result: Fall time 1.56 s; impact speed 15.34 m/s or 55.24 km/h.

Check: Independent energy check: $v^2=2gh=235.44\,\mathrm{m^2\,s^{-2}}$ . Also, $\frac{12\,\mathrm m}{1.56412\,\mathrm s}=7.67203\,\mathrm{m\,s^{-1}}$ , half the final speed.

Upward launch: is acceleration zero at the highest point?

Nonzero initial velocity

Given: Start at z₀ = 0 m with v₀ = +14 m/s, positive axis upward; g = 9.81 m/s².
Find: Find time to the peak, height gain and return time to the launch level.

Approach: now choose upward positive. Acceleration is −g. The peak occurs when vertical velocity is zero; return follows from equal initial and final height.
$\begin{aligned}v_z(t)&=v_0-gt\\z(t)-z_0&=v_0t-\frac12gt^2\\v_z(t_{\mathrm{peak}})&=0\\z(t_{\mathrm{return}})&=z_0\end{aligned}$
With upward positive, acceleration is −g.
$\begin{aligned}v(t)&=14-9.81t\\{} z(t)&=14t-\tfrac12\cdot9.81t^2\end{aligned}$
At the peak v = 0, not a = 0.
$t_{\mathrm{peak}}=\frac{14}{9.81}=1.42712\ldots\,\mathrm s$
Calculate peak height without premature rounding.
$\Delta z_{\max}=\frac{14^2}{2\cdot9.81}=9.98981\ldots\,\mathrm m$
z = 0 has the launch root and the later return root.
$\begin{aligned}t(14-4.905t)&=0\\{} t_{\mathrm{return}}&=\frac{28}{9.81}=2.85423\ldots\,\mathrm s\end{aligned}$

Result: Peak at 1.43 s with a 9.99 m height gain; return to launch level at 2.85 s with v = −14 m/s.

Check: Ascent and descent take equal time in this model because initial and final levels coincide. Acceleration remains −9.81 m/s² at the peak.

Do heavier objects fall faster? An energy check in the ideal model

mgh=\frac12mv^2\quad\Longrightarrow\quad v^2=2gh\quad(m>0)

In the ideal model, lost gravitational potential energy becomes kinetic energy.[3]

Interpreting a drop experiment or well measurement

T=t_{\mathrm{fall}}+t_{\mathrm{sound}}=\sqrt{\frac{2h}{g}}+\frac hc

h/c is in seconds; c must be specified as a separate model assumption.

Drag, terminal speed and varying gravity

Five solved falling exercises

Calculate · Foundation

An object falls from rest for 1.2 s. Find distance and speed using g = 9.81 m/s².

Show hint

Square t for distance, not for speed.

Learning goal: Find fall distance and velocity from rest.

Show solution and calculation

Approach: release from rest gives h = ½gt² and downward speed v = gt.
At 1.2 s the distance fallen is 7.0632 m and speed is 11.772 m/s.

Check: average speed over the drop is half the final speed; multiplying it by 1.2 s gives the same distance. Earlier contact with the ground would terminate the model.

\begin{aligned}h&=\frac12(9.81\,\mathrm{m/s^2})(1.2\,\mathrm s)^2=7.0632\,\mathrm m\\v&=(9.81\,\mathrm{m/s^2})(1.2\,\mathrm s)=11.772\,\mathrm{m/s}\end{aligned}

[1][2]

Rearrange · Foundation

How long does a drop from rest through 5 m take without air resistance?

Show hint

Use the positive time root.

Learning goal: Infer fall time from drop height.

Show solution and calculation

Approach: solve h = ½gt² for t.
The positive root represents time since release.
A five-metre drop takes about 1.01 s.

Check: substitution into ½gt² returns 5 m. The quantity beneath the root has units of seconds squared, so taking its root yields seconds.

\begin{aligned}t&=\sqrt{\frac{2h}{g}}\\&=\sqrt{\frac{2(5\,\mathrm m)}{9.81\,\mathrm{m/s^2}}}\\&\approx1.010\,\mathrm s\end{aligned}

[1][2]

Transfer · Application

An object is thrown downward at 3 m/s. How far does it fall in 0.8 s?

Show hint

With downward positive, v₀ and a are both positive.

Learning goal: Include nonzero initial throwing velocity.

Show solution and calculation

Approach: the initial downward velocity contributes v₀t.
Add it to the acceleration contribution ½gt².
At 0.8 s these contributions are 2.4 m and 3.1392 m, giving 5.5392 m.

Check: the result must exceed the distance of a drop from rest lasting the same time, since downward motion already exists at the start.

\begin{aligned}h&=v_0t+\frac12gt^2\\&=(3\,\mathrm{m/s})(0.8\,\mathrm s)+\frac12(9.81\,\mathrm{m/s^2})(0.8\,\mathrm s)^2\\&=2.4\,\mathrm m+3.1392\,\mathrm m=5.5392\,\mathrm m\end{aligned}

[1][2]

Interpret · Foundation

An object is thrown vertically upward at 9.81 m/s in vacuum, with constant g=9.81 m/s². Positive is upward. State time, velocity and acceleration at the apex. Is the ascending phase also free fall?

Show hint

Free fall describes the acting force, not instantaneous motion direction.

Learning goal: Distinguish zero speed at the apex from zero acceleration.

Show solution and calculation

Approach: $v=v_0-gt$ reaches zero after 1 s.
Acceleration remains −9.81 m/s²; gravity acts both before and after the apex.
Ascending motion is also free fall once the throwing hand loses contact.

Check: velocity is positive just before the apex and negative just after. Without continued acceleration, reversal could not occur.

\begin{aligned}t_{\mathrm{apex}}&=\frac{v_0}{g}=1\,\mathrm s\\v_{\mathrm{apex}}&=0\\a&=-9.81\,\mathrm{m/s^2}\end{aligned}

[1][2]

Error analysis · Application

A stone is dropped from rest into a vertical shaft. Time from release to hearing impact is 2.00 s. c=340 m/s, g=9.81 m/s²; neglect drag and reaction time. A solution gives h=19.62 m. Is that too large or too small? Find corrected depth.

Show hint

Fall time is less than 2 s because sound also needs time.

Learning goal: Model measured delay as fall time plus sound travel time.

Show solution and calculation

Approach: $2\,\mathrm s=t_f+\frac hc$ with $h=\frac12gt_f^2$ .
The positive solution gives tf≈1.94540 s and h≈18.5634 m.
The original 19.62 m is too large because the entire delay was counted as fall time.

Check: sound travel of about 0.05460 s completes 2.00 s. This assumes sound traverses the same vertical distance at the supplied constant c.

\begin{aligned}T&=t_f+\frac{gt_f^2}{2c}\\t_f&=\frac{2T}{1+\sqrt{1+\frac{2gT}{c}}}\\h&=\frac12gt_f^2\end{aligned}

[1][2]

What to take away

First identify whether height means distance fallen or height above the ground, and whether initial velocity is zero.
Weight and acceleration point downward even at the peak. Only velocity is momentarily zero there for a vertical throw.
The ideal equations describe motion before contact. Air forces and sound travel time require separate modelling when relevant.

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Distinguished free fall from a drop from rest directly; made fall time, impact speed, the highest point and mass accessible through appropriate headings. Linked the general displacement law, potential energy and momentum change during contact.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added prerequisites and an intuitive introduction; clarified the derivation, worked-example approaches and exercise solutions, and added a summary.

Cite this reference page

Quanta. Free Fall. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/freier-fall

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Frequently asked questions about Free Fall

How do you calculate the fall height with

h = \frac{1}{2}gt^{2}

Do heavy bodies really not fall faster than light ones?+

How do you calculate the fall time from the height?+

At what speed does a falling body hit the ground?+

What changes for a fall with initial velocity or on other planets?+

Free fall: vertical motion under gravity

Variables & units

Free fall calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Falling and projectile motion

Worked example

Applications

How do you calculate with Free Fall?

Getting an intuitive picture

Which free-fall equations apply to a drop from rest and a vertical launch?

The force remains downward

Fall velocity and fall time from the motion model

Distance fallen and height above ground

Fall time and impact speed: a twelve-metre example

Time and speed from drop height

Upward launch: is acceleration zero at the highest point?

Nonzero initial velocity

Do heavier objects fall faster? An energy check in the ideal model

Interpreting a drop experiment or well measurement

Drag, terminal speed and varying gravity

Five solved falling exercises

An object falls from rest for 1.2 s. Find distance and speed using g = 9.81 m/s2.

How long does a drop from rest through 5 m take without air resistance?

An object is thrown downward at 3 m/s. How far does it fall in 0.8 s?

An object is thrown vertically upward at 9.81 m/s in vacuum, with constant g=9.81 m/s2. Positive is upward. State time, velocity and acceleration at the apex. Is the ascending phase also free fall?

A stone is dropped from rest into a vertical shaft. Time from release to hearing impact is 2.00 s. c=340 m/s, g=9.81 m/s2; neglect drag and reaction time. A solution gives h=19.62 m. Is that too large or too small? Find corrected depth.

What to take away

Sources for this explanation

Frequently asked questions about Free Fall

Subject references and further reading

More formulas: Mechanics

Retain Free Fall for exams

Free fall: vertical motion under gravity

Variables & units

Free fall calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Falling and projectile motion

Worked example

Applications

How do you calculate with Free Fall?

Getting an intuitive picture

Which free-fall equations apply to a drop from rest and a vertical launch?

The force remains downward

Fall velocity and fall time from the motion model

Distance fallen and height above ground

Fall time and impact speed: a twelve-metre example

Time and speed from drop height

Upward launch: is acceleration zero at the highest point?

Nonzero initial velocity

Do heavier objects fall faster? An energy check in the ideal model

Interpreting a drop experiment or well measurement

Drag, terminal speed and varying gravity

Five solved falling exercises

An object falls from rest for 1.2 s. Find distance and speed using g = 9.81 m/s2.

How long does a drop from rest through 5 m take without air resistance?

An object is thrown downward at 3 m/s. How far does it fall in 0.8 s?

An object is thrown vertically upward at 9.81 m/s in vacuum, with constant g=9.81 m/s2. Positive is upward. State time, velocity and acceleration at the apex. Is the ascending phase also free fall?

A stone is dropped from rest into a vertical shaft. Time from release to hearing impact is 2.00 s. c=340 m/s, g=9.81 m/s2; neglect drag and reaction time. A solution gives h=19.62 m. Is that too large or too small? Find corrected depth.

What to take away

Sources for this explanation

Frequently asked questions about Free Fall

Subject references and further reading

More formulas: Mechanics

Retain Free Fall for exams

An object falls from rest for 1.2 s. Find distance and speed using g = 9.81 m/s².

An object is thrown vertically upward at 9.81 m/s in vacuum, with constant g=9.81 m/s². Positive is upward. State time, velocity and acceleration at the apex. Is the ascending phase also free fall?

A stone is dropped from rest into a vertical shaft. Time from release to hearing impact is 2.00 s. c=340 m/s, g=9.81 m/s²; neglect drag and reaction time. A solution gives h=19.62 m. Is that too large or too small? Find corrected depth.

An object falls from rest for 1.2 s. Find distance and speed using g = 9.81 m/s².

An object is thrown vertically upward at 9.81 m/s in vacuum, with constant g=9.81 m/s². Positive is upward. State time, velocity and acceleration at the apex. Is the ascending phase also free fall?

A stone is dropped from rest into a vertical shaft. Time from release to hearing impact is 2.00 s. c=340 m/s, g=9.81 m/s²; neglect drag and reaction time. A solution gives h=19.62 m. Is that too large or too small? Find corrected depth.