Free Fall — padho.wiki

In short

Free fall is motion under gravity alone, with a constant downward acceleration g \approx 9.8 m/s². A dropped object falls a distance h = \tfrac{1}{2}g t^2 and hits the ground at speed v = gt. An object thrown straight up at speed u reaches maximum height u^2/(2g) at time u/g, then falls back symmetrically — total flight time is 2u/g. In vacuum, all objects fall at the same rate regardless of mass.

Stand on the edge of the Qutub Minar observation deck, 73 metres above the ground, and hold a one-rupee coin over the railing. Let go. The coin does not float. It does not drift sideways. It drops — straight down, faster and faster, accelerating every tenth of a second until it strikes the ground about 3.9 seconds later at nearly 140 km/h.

Now picture a second experiment. Stand on the same deck and throw a cricket ball straight up as hard as you can. The ball climbs, slows, stops for a frozen instant at the very top, then falls back past you, accelerating downward all the way to the ground. Two very different motions — one starts from rest, the other starts fast — yet both are governed by exactly the same rule: the only force acting is gravity, and gravity pulls everything downward at the same constant rate.

This is free fall: motion where gravity is the only force. No air resistance, no engine, no string — just the pull of the Earth. And because gravity gives every object the same acceleration, free fall is the purest example of uniformly accelerated motion you will ever meet.

Why all objects fall at the same rate in vacuum

Before you touch a single equation, consider the deepest fact about free fall — the one that took humanity centuries to discover.

Drop a cricket ball and a sheet of paper from the same height. The ball hits the ground first. This is so obvious that for nearly two thousand years, people believed heavy objects fall faster. But the paper falls slowly because air pushes back on it, not because gravity pulls it less. If you crumple the paper into a tight ball and drop it again, it falls almost as fast as the cricket ball. Remove the air entirely — put both objects inside a vacuum chamber — and they hit the ground at exactly the same instant.

The fact: In the absence of air resistance, every object near Earth's surface — a coin, a cricket ball, a feather, an ISRO satellite casing — accelerates downward at the same rate, regardless of mass, size, or shape. This rate is called g and is approximately 9.8 m/s².

Why does mass not matter? Here is the physical argument. A heavier object does feel a stronger gravitational pull — the force of gravity on a 2 kg ball is twice the force on a 1 kg ball. But the heavier object also resists acceleration more — it has twice the inertia. The two effects cancel exactly:

F = mg \qquad \text{and} \qquad F = ma

mg = ma

a = g

Why: the m on both sides is the same mass (this is a deep fact — gravitational mass equals inertial mass). It cancels, leaving the acceleration independent of mass.

Every object in free fall has the same acceleration: g \approx 9.8 m/s², directed straight down. This is the starting point for everything that follows.

Free fall as uniformly accelerated motion

Since g is constant (near Earth's surface), free fall is simply a special case of uniformly accelerated motion with a = g. You already know the three kinematic equations from uniformly accelerated motion. In free fall, they become:

General form	Free-fall form (downward positive)
v = u + at	v = u + gt
s = ut + \tfrac{1}{2}at^2	h = ut + \tfrac{1}{2}g t^2
v^2 = u^2 + 2as	v^2 = u^2 + 2gh

Here h is the distance fallen (measured downward from the release point), u is the initial speed (positive downward), and v is the speed after falling through height h or after time t.

Assumptions: Neglect air resistance. For a dense, compact object like a coin or a cricket ball over a few seconds of fall, the drag force is tiny compared to gravity, so this is an excellent approximation. (For a feather or a parachute, air resistance dominates and these equations no longer apply.)

Dropping from rest — the simplest case

When you release an object from rest, u = 0. The three equations simplify beautifully.

Distance fallen after time t:

Start with s = ut + \tfrac{1}{2}at^2. Set u = 0 and a = g:

h = \tfrac{1}{2}g t^2 \tag{1}

Why: with no initial velocity, the entire distance comes from the acceleration term. The t^2 tells you the distance grows quadratically — the object covers more and more ground each second because it is speeding up.

Speed after time t:

Start with v = u + at. Set u = 0:

v = gt \tag{2}

Why: the speed increases linearly with time. After 1 second, the speed is 9.8 m/s. After 2 seconds, 19.6 m/s. After 3 seconds, 29.4 m/s. The object gains 9.8 m/s of speed every single second.

Speed after falling height h:

Start with v^2 = u^2 + 2as. Set u = 0:

v^2 = 2gh \tag{3}

v = \sqrt{2gh}

Why: this eliminates time and directly connects the speed to how far the object has fallen. Drop a coin from 5 m and it hits the ground at \sqrt{2 \times 9.8 \times 5} = \sqrt{98} \approx 9.9 m/s — regardless of how long it took.

Here is a simulation of a coin dropped from rest, accelerating under gravity. Watch how the distance between successive positions grows — that is the acceleration at work.

A coin dropped from rest. Gravity accelerates it at $9.8$ m/s² — the growing gaps between ghost markers show the coin covering more distance each second. Click replay to watch again.

Look at the numbers. After 1 second, the coin has fallen 4.9 m. After 2 seconds, 19.6 m. After 3 seconds, 44.1 m. In the first second it fell 4.9 m; in the second second it fell 19.6 - 4.9 = 14.7 m; in the third second, 44.1 - 19.6 = 24.5 m. Each successive second, it falls farther — because it is faster.

Objects thrown upward

Now for the more interesting case: you throw a cricket ball straight up at speed u. Gravity still pulls downward at g, so the ball decelerates on the way up, stops momentarily at the peak, then accelerates back down.

The key insight is that gravity does not care which direction the ball is moving. The acceleration is g downward at every instant — on the way up, at the peak, and on the way down. There is no separate "going up" physics and "coming down" physics. It is all one continuous motion with one constant acceleration.

Let's choose upward as positive. Then a = -g (acceleration is downward, opposite to the positive direction), and the kinematic equations become:

v = u - gt \tag{4}

h = ut - \tfrac{1}{2}g t^2 \tag{5}

v^2 = u^2 - 2gh \tag{6}

Here h is the height above the launch point, u is the initial upward speed, and v is the velocity at time t (positive upward, negative downward).

Time to reach the peak

At the highest point, the ball's velocity is zero — it has stopped climbing and is about to start falling. Set v = 0 in equation (4):

0 = u - g\,t_{\text{peak}}

t_{\text{peak}} = \frac{u}{g} \tag{7}

Why: the ball loses g = 9.8 m/s of upward speed every second. Starting with speed u, it takes u/g seconds to lose all of it and stop. If u = 20 m/s, the time to the peak is 20/9.8 \approx 2.04 seconds.

Maximum height

The maximum height is the height at t = t_{\text{peak}}. Substitute t_{\text{peak}} = u/g into equation (5):

H = u \cdot \frac{u}{g} - \tfrac{1}{2}g\left(\frac{u}{g}\right)^2

H = \frac{u^2}{g} - \frac{u^2}{2g}

H = \frac{u^2}{2g} \tag{8}

Why: the first term u^2/g is the distance the ball would travel at constant speed u for time u/g. But it does not travel at constant speed — gravity slows it down, eating away exactly half of that distance. The net result is u^2/(2g).

You can also derive this directly from equation (6). At the peak, v = 0:

0 = u^2 - 2gH

H = \frac{u^2}{2g}

Why: this is the same answer arrived at differently — confirming the result by a second route. For u = 20 m/s: H = 400/19.6 \approx 20.4 m.

Total flight time

The ball returns to its launch height when h = 0 (it is back to where it started). Set h = 0 in equation (5):

0 = ut - \tfrac{1}{2}g t^2

0 = t\left(u - \tfrac{1}{2}g t\right)

This gives two solutions: t = 0 (the start) and:

t_{\text{total}} = \frac{2u}{g} \tag{9}

Why: the quadratic has two roots because the ball is at height zero twice — once when you throw it (start) and once when it returns (landing). The total flight time 2u/g is exactly twice the time to the peak, u/g. This is the first hint of symmetry.

The symmetry of ascent and descent

Compare the time going up (t_{\text{peak}} = u/g) to the time coming down (t_{\text{total}} - t_{\text{peak}} = 2u/g - u/g = u/g). They are identical. The ball spends exactly as much time going up as it does coming down.

But the symmetry goes deeper than time. Consider the speed at any height h on the way up. From equation (6):

v_{\text{up}}^2 = u^2 - 2gh

On the way down, when the ball passes through the same height h, it has been falling from the peak (height H = u^2/(2g)) through a distance H - h:

v_{\text{down}}^2 = 2g(H - h) = 2g\left(\frac{u^2}{2g} - h\right) = u^2 - 2gh

Why: the speed squared at height h is the same going up and coming down. The ball passes through every height at the same speed — the only difference is the direction. Going up, the velocity is positive (upward). Coming down, it is negative (downward). The magnitude is identical.

This means if you throw a ball up at 20 m/s, it returns to your hand at 20 m/s (downward). Gravity takes away speed on the way up and gives back exactly the same speed on the way down. The upward journey is a perfect mirror image of the downward journey.

The speed at each height is the same on the way up and the way down. The upward journey is a mirror image of the downward journey — only the direction reverses.

This symmetry is not a coincidence. It is a direct consequence of energy conservation. The kinetic energy at any height h is \tfrac{1}{2}mv^2 = \tfrac{1}{2}mu^2 - mgh — it depends only on h, not on whether the ball is going up or coming down.

Dropping vs throwing — different starts, same acceleration

Imagine two experiments from the top of Qutub Minar (73 m):

You drop a coin from rest (u = 0).
You throw a coin downward at 10 m/s (u = 10 m/s).

Both coins experience the same acceleration g = 9.8 m/s² downward. The only difference is the initial condition — how fast each starts. The thrown coin is always ahead (it had a head start in speed), but both accelerate at the same rate.

Different initial velocities, identical acceleration. The thrown coin is always faster and always ahead, but both accelerate at the same $9.8$ m/s² every second.

After 1 second: the dropped coin has fallen 4.9 m and is moving at 9.8 m/s. The thrown coin has fallen 4.9 + 10 = 14.9 m and is moving at 10 + 9.8 = 19.8 m/s. The gap between them grows, but the acceleration — the rate at which speed increases — is identical for both.

This is the core lesson of free fall: gravity does not care how you started. Dropped, thrown down, thrown up — the acceleration is always g downward. The initial velocity u simply sets the starting condition for a universal equation.

Worked examples

Example 1: Dropping a coin from Qutub Minar

A coin is dropped from rest from the top of Qutub Minar, 73 m above the ground. Neglecting air resistance, find (a) the time it takes to hit the ground, and (b) the speed at impact.

A coin dropped from the top of Qutub Minar (73 m). The widening gaps between ghost markers show the acceleration — the coin falls farther in each successive second. Click replay to watch again.

Step 1. Identify the knowns.

h = 73 m, u = 0 (dropped from rest), a = g = 9.8 m/s².

Why: "dropped from rest" means the initial velocity is zero. The coin starts stationary and gravity does all the work.

Step 2. Find the time to hit the ground. Use h = \tfrac{1}{2}g t^2:

73 = \tfrac{1}{2}(9.8)\,t^2

t^2 = \frac{73}{4.9} = 14.90

t = \sqrt{14.90} \approx 3.86 \text{ s}

Why: rearranging h = \tfrac{1}{2}g t^2 for t gives t = \sqrt{2h/g}. The coin takes just under 4 seconds to fall 73 metres — a surprisingly short time for such a tall structure.

Step 3. Find the impact speed. Use v = gt:

v = 9.8 \times 3.86 \approx 37.8 \text{ m/s}

Why: the speed grows linearly with time. Converting to km/h: 37.8 \times 3.6 \approx 136 km/h — faster than a fast bowler's delivery.

Verify with the other formula:

v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 73} = \sqrt{1430.8} \approx 37.8 \text{ m/s} \checkmark

Result: The coin takes approximately 3.86 seconds to reach the ground and hits at about 37.8 m/s (\approx 136 km/h).

What this shows: Even a small, light coin reaches dangerous speeds when dropped from 73 metres. The height-to-speed formula v = \sqrt{2gh} gives the impact speed directly — no need to find the time first.

Example 2: Cricket ball thrown straight up

A cricket ball is thrown straight up at 20 m/s from ground level. Find (a) the maximum height, (b) the time to reach the peak, and (c) the total time in the air.

A cricket ball launched straight up at 20 m/s. It rises, slows, momentarily stops at 20.4 m, then falls back symmetrically. The ghost marker at $t = 2.04$ s sits at the peak. Click replay to watch again.

Step 1. Identify the knowns.

u = 20 m/s (upward), a = -g = -9.8 m/s² (gravity opposes the initial velocity).

Why: taking upward as positive, the acceleration is negative because gravity pulls downward. The ball decelerates on the way up.

Step 2. Find the maximum height. At the peak, v = 0. Use v^2 = u^2 - 2gH:

0 = (20)^2 - 2(9.8)\,H

H = \frac{400}{19.6} = 20.4 \text{ m}

Why: all the kinetic energy \tfrac{1}{2}mu^2 at launch converts to potential energy mgH at the peak. Cancelling m gives H = u^2/(2g) — a formula worth memorising.

Step 3. Find the time to the peak. Use v = u - gt_{\text{peak}} with v = 0:

0 = 20 - 9.8\,t_{\text{peak}}

t_{\text{peak}} = \frac{20}{9.8} \approx 2.04 \text{ s}

Why: the ball loses 9.8 m/s of speed each second. Starting at 20 m/s, it takes 20/9.8 \approx 2.04 seconds to reach zero speed.

Step 4. Find the total time in the air. By symmetry, the descent takes the same time as the ascent:

t_{\text{total}} = 2\,t_{\text{peak}} = 2 \times 2.04 = 4.08 \text{ s}

Why: the ball returns to the ground at the same speed it left (20 m/s, now downward). The upward journey is a mirror of the downward journey, so each takes u/g seconds.

Verify: Use h = ut - \tfrac{1}{2}g t^2 at t = 4.08 s:

h = 20 \times 4.08 - 4.9 \times (4.08)^2 = 81.6 - 81.6 = 0 \checkmark

The ball is back at ground level.

Result: Maximum height \approx 20.4 m. Time to peak \approx 2.04 s. Total flight time \approx 4.08 s.

What this shows: The three quantities — peak height, time to peak, total time — are all determined by a single number: the initial speed u. Double the launch speed and you quadruple the height (since H \propto u^2) but only double the time (since t \propto u).

Common confusions

"Heavy objects fall faster." In air, a cricket ball does fall faster than a feather — but that is because air resistance slows the feather more, not because gravity pulls the ball harder. In vacuum, both fall at exactly 9.8 m/s². The Apollo 15 astronaut David Scott dropped a hammer and a feather on the Moon (no atmosphere) — they hit the ground at the same instant. The same would happen in a vacuum chamber on Earth.
"Acceleration is zero at the top of an upward throw." This is a very common mistake. At the peak, the velocity is zero, but the acceleration is still g = 9.8 m/s² downward — gravity does not switch off when the ball stops momentarily. If the acceleration were zero at the top, the ball would float there forever.
"Free fall means falling down." Not necessarily. An object thrown upward is also in free fall the moment it leaves your hand — gravity is the only force acting on it, whether it is going up or coming down. "Free fall" means "gravity only," not "heading toward the ground."
"The equations change when the ball is going up vs coming down." The same equation h = ut - \tfrac{1}{2}g t^2 describes the entire motion — ascent, peak, and descent. You do not need separate equations for the two halves. The mathematics handles the direction change automatically through the signs.
"g is always 9.8 m/s²." Not exactly. g varies slightly with latitude (about 9.78 m/s² at the equator, 9.83 m/s² at the poles) and altitude (weaker at higher altitudes). For most problems, 9.8 m/s² (or sometimes 10 m/s² for quick estimates) is a fine approximation.

If you came here to understand free fall, use the formulas, and solve problems involving dropped and thrown objects, you have everything you need. What follows explores the Diwali rocket scenario, the effect of air resistance, and the connection to projectile motion.

The Diwali rocket — upward launch with thrust

A Diwali rocket is not in free fall while its fuel is burning — the rocket engine exerts an upward thrust in addition to gravity. But the moment the fuel runs out, the rocket is in free fall. If the rocket is moving upward at speed v_0 when the fuel is exhausted at height h_0, you can treat everything after that as an upward throw from height h_0 at speed v_0.

The maximum additional height above the burnout point is:

\Delta h = \frac{v_0^2}{2g}

And the total maximum height is h_0 + v_0^2/(2g). The rocket then falls freely from that height back to the ground.

This is exactly how ISRO analyses the trajectory of a sounding rocket after burnout — the powered phase is handled with the thrust equation, and the unpowered coasting phase is pure free fall.

Air resistance and terminal velocity

Real falling objects do experience air resistance. The drag force depends on the object's speed, cross-sectional area, and shape:

F_{\text{drag}} = \tfrac{1}{2}\rho\,C_d\,A\,v^2

where \rho is the air density, C_d is the drag coefficient, A is the cross-sectional area, and v is the speed.

As an object falls faster, the drag force increases (it goes as v^2). At some speed, the drag force equals the gravitational force mg, and the net force becomes zero. The object stops accelerating and falls at a constant speed called the terminal velocity:

v_{\text{terminal}} = \sqrt{\frac{2mg}{\rho\,C_d\,A}}

Why: set F_{\text{drag}} = mg and solve for v. A heavier object (larger m) or a more streamlined one (smaller C_d A) has a higher terminal velocity. A skydiver has a terminal velocity of about 55 m/s (200 km/h) in the spread-eagle position, but about 80 m/s (290 km/h) in a head-down dive — same mass, different C_d A.

For a cricket ball (m \approx 0.16 kg, C_d \approx 0.5, A \approx 0.004 m², \rho \approx 1.2 kg/m³):

v_{\text{terminal}} = \sqrt{\frac{2 \times 0.16 \times 9.8}{1.2 \times 0.5 \times 0.004}} \approx \sqrt{653} \approx 36 \text{ m/s}

This means a cricket ball dropped from the Qutub Minar (where the free-fall equation predicts 37.8 m/s impact speed) would actually be close to its terminal velocity — air resistance would have a noticeable but modest effect. For a coin, the terminal velocity is much higher, so the free-fall approximation is excellent.

The connection to projectile motion

Free fall is one-dimensional projectile motion. When you add a horizontal velocity component, the vertical motion is unchanged — the ball still accelerates downward at g — but now there is also constant horizontal motion. The result is a parabolic trajectory.

The key insight from free fall carries over directly: the vertical and horizontal motions are independent. The horizontal velocity does not affect how fast the object falls. This is explored in detail in Projectile Motion Fundamentals.

Measuring g — the simple drop experiment

You can measure g with a ruler and a stopwatch. Drop a dense object (a steel ball or a coin) from a measured height h and time the fall. Then:

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

The challenge is timing accuracy — for a 2 m drop, the fall takes only 0.64 seconds, and a 0.05 s error in timing gives a 15\% error in g. Better methods use electronic timers triggered by the ball passing through light gates, or measure the spacing of dots on a tape pulled through a ticker timer.

The accepted value g = 9.81 m/s² at sea level (standard gravity) has been measured to extraordinary precision using falling corner-cube reflectors and laser interferometry.

Where this leads next

Graphs of Motion — how to read and draw position-time, velocity-time, and acceleration-time graphs for free fall and other uniformly accelerated motions.
Projectile Motion Fundamentals — what happens when you add a horizontal velocity to free fall, producing the parabolic trajectory.
Acceleration — the general concept of acceleration, of which g is one constant example.
Uniformly Accelerated Motion — the three kinematic equations in their general form, which free fall is a special case of.