An Introduction to Free Fall Without Air Resistance

Under Taiwan’s current 108 Curriculum Guidelines, students first encounter free fall in ninth-grade science when uniformly accelerated motion is introduced. The topic appears again in first-year high-school basic physics and in the kinematics unit of second-year mechanics.

Free fall keeps showing up after that as well—in Newtonian mechanics, energy problems, and even the motion of charged particles between parallel plates in electrostatics. It is basically everywhere XD. That is physics for you: once you learn something, it keeps coming back later, combined with earlier ideas and turned into new exam questions.

Definition of Free Fall

Free fall is defined here as one-dimensional motion in which an object starts at rest relative to the observer and moves only under the gravitational force near a planet’s surface[1][2].

Analyzing Free Fall Near Earth’s Surface

Near sea level at a latitude of 45°, gravitational acceleration is approximately constant at 9.8 m/s2[1]. For convenience, I will use g to represent the magnitude of gravitational acceleration from this point on.

We can therefore treat free fall near Earth’s surface as one-dimensional uniformly accelerated motion. Because the initial speed is zero and the gravitational acceleration g points toward Earth’s center, let the initial speed be v0, the final speed be v, the falling distance be h, and the elapsed time be t. If we define the direction toward Earth’s center—that is, “downward” in everyday language—as positive and set the release point to zero, the familiar uniformly accelerated-motion equations can be rewritten as follows:

Free-fall velocity: v = v₀ + at; starting from rest, v = gt with downward positive.
Displacement: ΔX = v₀t + ½at²; falling from rest through height h gives h = ½gt².
Speed and displacement: v² = v₀² + 2aΔX; falling from rest gives v² = 2gh.

Combining the basic properties of uniformly accelerated motion with the conditions of free fall gives the following vertical position–time (x–t or h–t), velocity–time (v–t), and acceleration–time (a–t) graphs:

For uniformly accelerated motion, the x–t graph is quadratic. Because free fall starts at the release point and the speed increases, the curve is an upward-opening parabola through the origin.
For uniformly accelerated motion, the x–t graph is quadratic. Because free fall starts at the release point and the speed increases, the curve is an upward-opening parabola through the origin.
For uniformly accelerated motion, the v–t graph is a straight line. Because free fall begins from rest, the line passes through the origin.
For uniformly accelerated motion, the v–t graph is a straight line. Because free fall begins from rest, the line passes through the origin.
With downward defined as positive, the gravitational acceleration is the constant value g, so the a–t graph is a horizontal line.
With downward defined as positive, the gravitational acceleration is the constant value g, so the a–t graph is a horizontal line.

Why Raindrops Raise a Free-Fall Problem

From the free-fall analysis above, if air resistance is ignored, an object falling from a height is acted on by the gravitational acceleration g throughout its motion. The uniformly accelerated-motion equations therefore give the final speed after an object falls through a height h:

Without air resistance, impact speed follows v² = 2gh, so v = √(2gh).

With that idea in hand, we can estimate the final speed of a raindrop falling from high in the sky.

Before doing that, however, we need a basic idea of how clouds are classified and roughly how high they are. Otherwise, we do not know what falling distance is reasonable for the raindrop we want to discuss.

The World Meteorological Organization (WMO), following the cloud-classification tradition that grew from meteorologist Luke Howard’s 1803 Essay on the Modification of Clouds, divides clouds into ten genera according to their shape, composition, and formation. These genera are commonly grouped into high clouds at roughly 6,000–7,000 m above the ground, middle clouds at roughly 2,000–6,000 m, and low clouds below roughly 2,000 m[3]. Vertically developed clouds span several of these height ranges and often produce short but very heavy rainfall[4].

Basic cloud-classification chart provided by the World Meteorological Organization [3].
Basic cloud-classification chart provided by the World Meteorological Organization [3].

Under the WMO classification and Howard’s terminology, the main rain-producing clouds are cumulonimbus and nimbostratus. The cloud base of nimbostratus varies with the conditions. For this worked example, I will assume a raindrop begins falling from 1,200 m above the ground. That is a model assumption, not a fixed cloud-base height for all nimbostratus clouds.

If a raindrop falls from high above and we substitute that height into the free-fall equation, we obtain the following theoretical final speed:

Idealized no-drag fall from 1,200 meters with g = 9.8 m/s² gives about 153 m/s.

The calculation says that if a raindrop fell from that height with no air resistance at all, it would reach the ground at about 153 m/s. Does that number still feel a little abstract? Let us do one more calculation.

A 2009 study[5] showed that small falling raindrops are nearly spherical, while larger drops become increasingly flattened by their interaction with the air. As a drop grows, aerodynamic effects become stronger until sufficiently large drops break apart; this limits the largest raindrops to roughly 6 mm in diameter. Another study of raindrop-size distributions[6] found that most drops are about 0.5–4 mm in diameter, corresponding to radii of about 0.25–2 mm.

Illustration of how raindrops of different sizes deform as they interact with the air [7].
Illustration of how raindrops of different sizes deform as they interact with the air [7].

Ignoring differences in chemical composition and density changes during the fall, let us simply assume that each raindrop is water with a density of 1 g/cm3. If we further model the drop as a uniform sphere with the largest representative radius of 2 mm, the density–mass–volume relationship gives the mass of one raindrop as follows:

A spherical water drop of radius 2 mm and density 1,000 kg/m³ has mass about 3.3 × 10⁻⁵ kg.

Next, we can use Newton’s second law together with impulse and momentum to estimate the impact force of one raindrop. Imagine that you are standing in the rain, that the drop is in contact with your head for 0.001 s, and that it comes completely to rest on impact. Two forces contribute to the change in the drop’s velocity: gravity—an easy one to forget when setting up the problem XD—and the normal force from your head.

By Newton’s third law, the force your head exerts on the raindrop and the force the raindrop exerts on your head form an action–reaction pair. The latter is the impact force we actually want to estimate.

Using N for the force exerted by your head on the raindrop, the assumptions above give:

Idealized impact: N = m(153 m/s)/0.001 s + mg, about 5.049 N or 0.52 kgw, with no air drag.

Although gravity should technically be included when solving for the normal force N, the raindrop itself is so light that the gravitational term is negligible compared with the impact term. The resulting force is about 0.52 kgw. Still does not feel intuitive? In force magnitude alone, that is roughly the weight of a full 500 mL bottle of water; it is the average force during an assumed 0.001-second impact in a no-air-resistance model, not the sensation of a bottle pressing steadily on you.

That estimate is for just one raindrop. If we follow this idealized model, many drops arriving in succession seem as though they should feel dramatic; we cannot simply add brief impacts as if bottles of water were pressing on you at the same time.

Wait a minute. That does not sound anything like what ordinary rain actually feels like. So where did the model go wrong?

Raindrops with Air Resistance and Terminal Velocity

Rain clouds exist in the troposphere, which is full of air molecules. As a raindrop moves through this fluid made of air, it experiences not only gravity but also an additional force: air resistance, or drag.

In fluid dynamics, an object moving through a fluid experiences a resistive force from the fluid in the direction opposite to its motion. This kind of resistance occurs between fluid layers and between fluids and solids. Unlike the solid-on-solid friction usually introduced earlier in school physics, drag on an object moving through a fluid depends on the object’s speed[8][9].

The drag force depends on the properties of the fluid and on the object’s size, shape, and other characteristics. A common way to describe it is the drag equation[8]:

Quadratic drag equation: F_D = ½ρC_DAv².

Here, ρ is the fluid density (for motion in air, the average density of air), A is the object’s effective projected area, v is its speed relative to the fluid, and CD is the dimensionless drag coefficient. In general, the drag coefficient depends on the object’s shape and on the Reynolds number.

The Reynolds number is a dimensionless quantity in fluid dynamics that compares inertial effects with viscous effects and helps predict the flow regime. Different problems may use slightly different forms, but the expression generally involves fluid density, viscosity, flow speed, and a characteristic length or size. A basic form is[10]:

Reynolds number: Re = ρvD/μ.

Here, ρ is the fluid density, v is the average flow speed, D is the characteristic length, and μ is the dynamic viscosity. At low Reynolds numbers, the flow tends to be laminar: fluid particles move smoothly in layers, with adjacent layers sliding past one another like stacked cards and with little mixing or vortex formation[11]. At high Reynolds numbers, the flow tends to become turbulent: velocity and pressure vary irregularly, the neat parallel layers disappear, and mixing becomes common[12].

Panel (a) shows laminar streamlines, while panel (b) shows turbulent streamlines [13].
Panel (a) shows laminar streamlines, while panel (b) shows turbulent streamlines [13].
Viscosity represents a fluid’s resistance to deformation and flow. Higher-viscosity fluids flow less easily than lower-viscosity fluids. In this simulation, the fluid on the left has lower viscosity and the fluid on the right has higher viscosity [14].
Viscosity represents a fluid’s resistance to deformation and flow. Higher-viscosity fluids flow less easily than lower-viscosity fluids. In this simulation, the fluid on the left has lower viscosity and the fluid on the right has higher viscosity [14].

Okay, that was a bit of a detour. Let us return to the drag equation. Experimental observations show that at relatively high Reynolds numbers—roughly speaking, denser fluids, faster motion, and lower viscosity—the drag coefficient can often be treated as nearly constant, so the drag force is approximately proportional to the square of speed. At low Reynolds numbers—roughly speaking, slower motion and stronger viscous influence—the drag coefficient scales approximately with the inverse of Reynolds number. Combining that behavior with the definition of Reynolds number gives an approximately linear relationship between drag and speed. The two forms can be written as follows:

At high Reynolds number, the drag coefficient C_D can be treated as approximately constant, so the drag equation remains close to its standard form and the drag force is proportional to the square of speed. The drag coefficient can be estimated from experimental data or tables.
At high Reynolds number, the drag coefficient C_D can be treated as approximately constant, so the drag equation remains close to its standard form and the drag force is proportional to the square of speed. The drag coefficient can be estimated from experimental data or tables.
At low Reynolds number, the drag coefficient depends on Reynolds number and is approximately inversely proportional to it. Because Reynolds number itself is proportional to speed, dimensional analysis reduces the drag law to a form that is linear in speed. The constant b depends on the object and the fluid.
At low Reynolds number, the drag coefficient depends on Reynolds number and is approximately inversely proportional to it. Because Reynolds number itself is proportional to speed, dimensional analysis reduces the drag law to a form that is linear in speed. The constant b depends on the object and the fluid.

Now let us return to the original raindrop problem. Earlier, by considering gravity alone, we estimated that a drop falling from our example height of 1,200 m could reach about 153 m/s. Once air resistance is included, drag always points opposite to the drop’s velocity. Let the raindrop have mass m and, for simplicity, be a sphere of radius R. Its force diagram is then:

A falling raindrop moves downward and experiences upward drag. The downward acceleration arrow depicts the stage before terminal velocity, when gravity exceeds drag; at terminal velocity acceleration is zero.
A falling raindrop moves downward and experiences upward drag. The downward acceleration arrow depicts the stage before terminal velocity, when gravity exceeds drag; at terminal velocity acceleration is zero.

In the drag equation, A is the effective projected area. For a spherical raindrop, that is the shaded circular cross-section in the figure above, which can be calculated from R using the area formula for a circle. We can now apply Newton’s second law to obtain the raindrop’s acceleration during the fall:

Newton's second law with downward positive: mg − F_D = ma.

For a high-Reynolds-number case, where the square-law drag approximation is appropriate, the expression becomes:

At higher Reynolds number with quadratic drag, a = g − ρC_DAv²/(2m).

For a low-Reynolds-number case, where the linear-drag approximation is appropriate, it becomes:

At lower Reynolds number with linear drag, a = g − bv/m.

Both simplified expressions tell the same qualitative story: as the raindrop speeds up, drag increases and the acceleration becomes smaller. Eventually the acceleration reaches zero. At that point the drop no longer speeds up and instead falls at a constant speed. That constant speed is the terminal velocity, vt. At terminal velocity, gravity and drag are in force balance, so we can solve for the terminal speed in the two Reynolds-number regimes:

Terminal velocity obtained for the high-Reynolds-number case.
Terminal velocity obtained for the high-Reynolds-number case.
Terminal velocity obtained for the low-Reynolds-number case.
Terminal velocity obtained for the low-Reynolds-number case.

For a rough estimate of a raindrop in the atmosphere, I will use the high-Reynolds-number model because it is closer to the situation of interest here. As before, assume a spherical raindrop with radius 2 mm and density 1,000 kg/m3. Based on the drop’s shape and an empirical estimate, take the drag coefficient to be about 0.6[15].

Calculating the actual terminal speed in the high-Reynolds-number model requires the fluid density. Since the fluid here is air, we can—very ideally, in the ideal-gas sense of “ideal” XD—use the ideal gas law to estimate air density at 1 atm and 20°C, then substitute that value into the terminal-velocity equation.

The ideal gas law is rearranged into an expression for gas density. P is pressure, V is volume, n is the amount of substance, T is absolute temperature, R is the ideal gas constant, M is molar mass, and ρ is gas density. All quantities are expressed in SI units.
The ideal gas law is rearranged into an expression for gas density. P is pressure, V is volume, n is the amount of substance, T is absolute temperature, R is the ideal gas constant, M is molar mass, and ρ is gas density. All quantities are expressed in SI units.

Substitute the atmospheric conditions used here—1 atm and 20°C—making sure every quantity is converted to SI units. The ideal-gas constant in this form is 8.314.

Ideal-gas relation PM = ρRT; at one atmosphere and 20°C, air density is about 1.204 kg/m³.

Combining the resulting air density with the other assumptions and substituting them into the high-Reynolds-number terminal-velocity equation gives:

Quadratic-drag terminal speed v_t = √[2mg/(ρ_air C_D A)]; this model gives about 8.52 m/s.

Once air resistance is included, this estimate gives a terminal speed of about 8.52 m/s—dramatically smaller than the 153 m/s predicted by the gravity-only model, roughly one twentieth as large. Under the same Newton’s-second-law reasoning, the normal force on your head during impact also falls to roughly one twentieth of the previous estimate. Now the result looks much more like the rain we actually experience.

Advanced Extension: Velocity as a Function of Time with Air Resistance

So far we have seen that a falling raindrop is acted on by both gravity and air resistance. Because drag depends on speed, the motion is not uniformly accelerated: the acceleration changes with time. Newton’s second law gives the acceleration relationship, and as speed increases, the acceleration decreases until it reaches zero and the drop continues downward at terminal velocity.

But what is the exact functional relationship between velocity and time for a raindrop falling through air? Answering that question requires some university-level mathematics—specifically, techniques for solving differential equations.

Low-Reynolds-Number Case: Velocity as a Function of Time

Starting from the equation above, rewrite it in differential-equation form:

Linear drag: mg − bv = m dv/dt, rearranged as −(m/b)dv = (v − mg/b)dt.

Now substitute a new variable to simplify the expression:

Let u = v − mg/b; separation and integration give u = u₀ exp(−bt/m).

Because the raindrop begins falling from rest, v=0 at t=0. Substituting that initial condition into u gives u0. Replacing the substituted variable with v again yields the final velocity–time relationship:

With linear drag and v(0) = 0, v(t) = (mg/b)[1 − exp(−bt/m)].

High-Reynolds-Number Case: Velocity as a Function of Time

Using the same idea, rewrite Newton’s second-law equation for the high-Reynolds-number drag model as a differential equation.

Quadratic drag: mg − ½ρC_DAv² = ma, hence dv/dt = g − ρC_DAv²/(2m).

Many quantities in this expression are constants—the raindrop’s projected area, weight, drag coefficient, gravitational acceleration, and so on. We can group most of them into a single constant c to simplify the equation. The dt side is straightforward; the dv side is more troublesome. Here I use an algebraic factorization approach to reduce the order of the expression involving v.

Let c = ρC_DA/(2m); separate as dv/(g − cv²) = dt and decompose into partial fractions.

After factorization, the expression becomes a sum of simpler terms that can be integrated separately, with an integration constant C added. Now apply the initial condition t=0. Because the raindrop starts from rest, v=0 at t=0 whether or not air resistance is included. Substitution gives C=0, so the constant term disappears from the later steps.

Integration yields 2√g t + C = (1/√c) ln[(√g + √c v)/(√g − √c v)]; v(0) = 0 makes C = 0.

After a sequence of algebraic simplifications, restore the constants that were previously grouped into c. The result is the velocity–time relationship for a raindrop under the high-Reynolds-number drag model:

Quadratic-drag speed v(t) = √(g/c)[exp(2√cg t) − 1]/[1 + exp(2√cg t)], tending to terminal speed.

To check whether the two velocity–time results make sense, we can use terminal velocity as a limiting test. If enough time passes, the acceleration approaches zero and the speed approaches a constant. In the extreme limit t→∞, both the high- and low-Reynolds-number expressions approach the terminal velocities derived earlier. That agreement is a useful consistency check.

Graphical Trends

We now have two velocity–time functions, but both look complicated enough that simply staring at the equations does not give much intuition. So I used GeoGebra to plot their general shapes. A real raindrop model contains many constants and assumptions; since the goal here is only to compare trends, I simplified those constants as much as possible and plotted only the overall behavior.

Because the goal is only to compare trends, the complicated constants are simplified to 1 while preserving the signs that control the direction and shape of the curves. The green line represents the low-Reynolds-number linear-drag case; the blue line represents the high-Reynolds-number quadratic-drag case.
Because the goal is only to compare trends, the complicated constants are simplified to 1 while preserving the signs that control the direction and shape of the curves. The green line represents the low-Reynolds-number linear-drag case; the blue line represents the high-Reynolds-number quadratic-drag case.
The blue and green curves follow broadly similar trends, but they are not identical.
The blue and green curves follow broadly similar trends, but they are not identical.

In both cases, the velocity approaches a limiting value after enough time. This is exactly what our earlier analysis of decreasing acceleration and terminal velocity predicted.

We can also look at how the tangent slope changes. Because these are velocity–time graphs, the tangent slope at each point is the acceleration. As time increases, the slope becomes smaller, which means the acceleration is decreasing. Again, that matches the conclusion from the Newtonian force analysis.

One caution is important: these plots do not use the full set of real environmental conditions or raindrop parameters. They therefore cannot tell us whether the high- or low-Reynolds-number case reaches terminal velocity first in a real situation. The time required to approach terminal velocity depends on the actual conditions.

Related Questions from Taiwan’s Major Entrance Exams

Air resistance and terminal velocity may look far beyond the formal junior- and senior-high-school curriculum after the analysis above. In practice, however, entrance exams have used these ideas repeatedly for many years. So the concepts are still worth knowing. Here are several representative historical questions.

【ROC 69 Joint University Entrance Exam】

A parachutist jumps from high altitude. Because of air resistance, there is a period before landing during which the parachutist descends at constant speed. If the terminal speed at landing is 10 m/s and the total mass of the person and equipment is 100 kg, approximately how much thermal energy is produced per second during this constant-speed descent?

【ROC 78 Joint University Entrance Exam】

A small stone of weight w is thrown vertically upward with initial speed V0. Air exerts a constant-magnitude resistive force f opposite to the stone’s direction of motion. Find: (1) the maximum height reached by the stone; and (2) the speed of the stone when it returns to the ground.

【ROC 79 Joint University Entrance Exam】

A falling raindrop experiences air resistance and eventually descends at constant speed before reaching the ground. A raindrop has mass 2.7×10−7 kg and falls at a constant speed of 20 m/s before landing. Assume that all heat generated by air resistance during this constant-speed motion is absorbed by the raindrop and that its mass remains unchanged. Approximately how many degrees does the raindrop’s temperature rise per second?

【ROC 101 GSAT】The question is rather long, so I’ll use a screenshot of the exam paper instead.

ROC 101 GSAT problem: two falling spheres with linear drag and speed–time curves approaching different terminal speeds.

【ROC 102 GSAT】This one is also fairly long, so I’ll use a screenshot of the exam paper.

ROC 102 GSAT problem: a falling object's speed–time graph levels near 6 m/s; questions concern distance and changing drag.

【ROC 107 GSAT】And yes, this one is long too, so I’m using the exam screenshot again. (Definitely not because I’m being lazy XD.)

ROC 107 GSAT problem: choose the speed–time curve that represents a sphere falling through air with drag.

Conclusion

If you made it all the way here, thank you for sticking with me. This article started from the free-fall material taught in junior- and senior-high-school physics, then moved into the real-world problem of rainfall and quietly smuggled in some university-level ideas along the way. The goal was to explain the relationship between rain and terminal velocity in reasonably plain language before returning to entrance-exam questions. Air resistance may receive only limited formal treatment in the curriculum, but exams use it often enough that it remains an important extension topic.

This article originally grew out of a short supplementary note in the high-school handouts I wrote early in my teaching career. Air-resistance questions appeared so often—every few years they seemed to come back and poke students again—that I wanted a small extra explanation. As I taught for more years and saw more problems, the old note started to feel inadequate. I had written it when I was only a second-year university student, when many ideas were not yet fully learned or fully thought through. Now that I am older and, hopefully, somewhat better at science, I figured it was time to fill in the missing pieces. Then I kept finding more and more material, and somehow the “small supplement” became this enormous article. Orz.

I also wanted the numerical estimates to be as grounded as possible. When I needed a cloud-base height or a typical raindrop size, I looked for meteorological references rather than making up convenient numbers. When a value was not directly available—such as the density of air at 1 atm and 20°C—I at least wanted the derivation to be explicit. I tried to make every major statement traceable to some basis or reference. I normally do not love citing Wikipedia in technical writing, but after finishing atmospheric dynamics and fluid mechanics I angrily threw away the textbooks because I thought I would never need them again XDDDD, so... here we are.

Writing this article made me revisit differential equations, chemistry, fluid mechanics, and several other topics. Fortunately, the old skills have not completely rusted away. Now that it is finally finished, I can only hope the next article does not spiral this far out of control again. (Famous last words.)

A slightly simplified version of this article was also published on PanSci. If you would like to support it there as well, you are very welcome to take a look :D

Why Don’t Raindrops Beat You Up on Rainy Days?

References

[1] Walker, J., Halliday, D., & Resnick, R. (2011). Fundamentals of physics. Hoboken, NJ: Wiley.

[2] Free fall - Wikipedia.

[3] World Meteorological Organization (WMO), Classifying Clouds.

[4] Central Weather Bureau (Taiwan), 天空的魔術師—千變萬化的雲朵 [The Magician of the Sky—Ever-Changing Clouds].

[5] Emmanuel Villermaux, Benjamin Bossa (2009). Single-drop fragmentation distribution of raindrops. Nature Physics. 5(9): 697–702.

[6] McFarquhar, Greg (2010). Raindrop Size Distribution and Evolution. Rainfall: State of the Science. Geophysical Monograph Series. Vol.191, p.49–60.

[7] Drop(liquid) - Wikipedia.

[8] Drag(physics) - Wikipedia.

[9] NASA(2010). What is drag? URL：https://web.archive.org/web/20100524003905/http://www.grc.nasa.gov/WWW/K-12/airplane/drag1.html

[10] Reynolds number - Wikipedia.

[11] Laminar flow - Wikipedia.

[12] Turbulence - Wikipedia.

[13] What is Laminar Flow? URL: https://www.simscale.com/docs/simwiki/cfd-computational-fluid-dynamics/what-is-laminar-flow/

[14] Viscosity - Wikipedia.

[15] Drag coefficient - Wikipedia.