Why doesn’t a falling raindrop keep speeding up?

Velocity is a derivative; acceleration is another

A falling object has a position $y(t)$. From the derivatives module, velocity is the rate at which position changes: $v = dy/dt$. Apply the same machine again and you get acceleration $a = dv/dt$ — the rate at which velocity changes. Newton’s second law packages this: net force per unit mass equals $dv/dt$. Once you can write down the forces and divide by mass, you have a recipe for predicting motion. The whole drama of this page sits in two terms on the right-hand side.

Two forces — gravity is constant, drag isn't

A falling raindrop has two forces acting on it. Gravity pulls down with constant magnitude $F_g = m·g$; per unit mass that’s just $g ≈ 9.8 m/s²$, regardless of how fast it’s moving. Drag pushes up against the motion, and unlike gravity it grows with speed. For small objects in air, a clean approximation is linear drag: $F_drag = k·v$ per unit mass. Slow raindrop → small drag. Fast raindrop → big drag. The two forces are vectors pointing opposite — exactly the simple-1D add-and-subtract case from the vectors module.

The net force per unit mass — what goes into $dv/dt$ — is the difference:

$dv/dt = g − k v$

Two terms. The first is constant. The second isn’t. That’s the whole equation.

What dv/dt = g − kv actually says

Read it left-to-right: at any instant, the rate at which velocity is changing equals gravity minus drag. The interesting thing is that drag depends on v itself: it’s part of the right-hand side of the equation that determines how v evolves. That’s what makes this a differential equation — the unknown function and its derivative both appear.

Three regimes, all readable from the formula alone:

• $v$ small (just dropped from rest): $kv$ is small, so $dv/dt ≈ g$ — full acceleration, just like vacuum free-fall.
• $v$ moderate: $kv$ is significant. $dv/dt = g − kv$ is positive but smaller. Velocity still grows, but more slowly.
• $v = g/k$ exactly: $kv = g$. $dv/dt = 0$. Velocity stops changing. This is terminal velocity .
• v > g/k (e.g., thrown downward fast): kv > g. dv/dt < 0. Velocity decreases — drag wins, the object slows down. Until it hits $v_t$ from above.

The widget’s blue curve is approaching the dashed green line from below; in either direction, $v_t$ is the destination.

import math

# Per-unit-mass equation of motion: dv/dt = g - k*v.
# Two competing forces, both per unit mass:
#   gravity  →  +g  (constant, pulls v upward in magnitude)
#   drag     →  -k*v  (proportional to current speed, opposes motion)
g = 9.8

def dvdt(v, k):
    return g - k * v

# At v = 0, drag is zero; gravity is unopposed; full acceleration.
# As v grows, drag grows proportionally; net force shrinks; acceleration
# shrinks. The shrink is the whole point — the system has a built-in
# governor.
[(v, round(dvdt(v, k=0.5), 2)) for v in (0, 5, 10, 15, 19.6, 25)]
# → [(0,    9.8),    free fall — nothing opposing
#    (5,    7.3),    drag has eaten 2.5 of g
#    (10,   4.8),
#    (15,   2.3),
#    (19.6, 0.0),    EQUILIBRIUM — v_t exactly
#    (25,  -2.7)]    above v_t: drag wins, deceleration

Terminal velocity = solving dv/dt = 0

The “destination” is just whatever value of $v$ makes $dv/dt = 0$. Set it and solve:

$g − k v_t = 0 → v_t = g / k$

One line of algebra. Larger drag (bigger $k$) gives smaller terminal velocity — a feather settles fast and slow; a bowling ball would take forever to reach $v_t$ in air (and in practice would hit the ground first). This is a equilibrium : the speed at which two competing influences exactly cancel. Gravity wants to add to $v$; drag wants to subtract from it; at $v_t$ they tie. Forces don’t disappear — they balance. That distinction is the page’s most common misunderstanding.

The formula $v_t = g/k$ isn’t a deep theorem; it’s a fixed-point reading. The same shape appears every time a quantity has competing influences that depend on the quantity itself: heating coffee in a cool room reaches the room temperature; a capacitor charging through a resistor reaches the supply voltage; a population with births and deaths proportional to size reaches a carrying capacity. Different physics, same fixed-point hunt.

# Solve dv/dt = 0 for terminal velocity. One line of algebra.
#   g - k*v_t = 0  →  v_t = g/k.
def terminal_velocity(k, g=9.8):
    return g / k

# Linear-drag terminal velocities for a few familiar k values.
[(name, k, round(terminal_velocity(k), 1))
 for name, k in (("feather", 6.0), ("raindrop", 1.6),
                 ("skydiver", 0.4), ("bowling ball", 0.05))]
# → [('feather',      6.0,  1.6),     reaches ~1.6 m/s in well under a second
#    ('raindrop',     1.6,  6.1),
#    ('skydiver',     0.4, 24.5),     classic ~50 mph free-fall figure
#    ('bowling ball', 0.05, 196.0)]   approaches very slowly, in practice
#                                      it'd hit the ground long before
# Real raindrops sit closer to 9 m/s because actual drag is closer to
# v² than v; the linear case is a teaching simplification.

The approach is asymptotic — and exponential

The closed-form solution from rest is

$v(t) = v_t · (1 − e^(−k t))$

The error from terminal — the gap $v_t − v(t)$ — is $v_t · e^(−k t)$: it shrinks exponentially. After $t = \ln 2 / k ≈ 0.693/k$ seconds, the error has halved. After $5/k$ seconds, the error is below 1% of $v_t$. The widget makes this readable: drag slider up (large $k$) pulls the curve hard to the asymptote in a fraction of a second; drag slider down (small $k$) lets it crawl up over many seconds.

Why exponential? Take the derivative of the error $ε(t) = v_t − v(t)$: $dε/dt = −dv/dt = −(g − kv) = −k ε$. The error’s rate of change is proportional to the error itself, with a negative sign. That’s the defining shape of exponential decay. Anything whose rate of change is proportional to itself ends up being an exponential — the logarithm module names this from the other direction. Linear-drag falling, capacitor charging, radioactive decay, Newton’s cooling: same equation, same answer.

# Closed-form solution from rest, by separation of variables.
#   v(t) = v_t * (1 - exp(-k*t))
# At t = 0, v = 0 (rest). As t → ∞, v → v_t. Half of v_t is reached at
# t = ln(2)/k ≈ 0.693/k. After 5 'time-constants' (5/k seconds), v is
# within 1% of v_t.
def v_of_t(t, k, g=9.8):
    return (g / k) * (1 - math.exp(-k * t))

[(t, round(v_of_t(t, k=0.4), 2)) for t in (0, 1, 3, 5, 10, 20)]
# → [(0,   0.0),
#    (1,   8.1),     a third of v_t
#    (3,  17.1),     two-thirds
#    (5,  21.2),     ~87% of v_t = 24.5
#    (10, 24.0),     ~98%
#    (20, 24.5)]     within 0.001
# Same shape as a charging capacitor, a heating cup of coffee, a leaky
# bucket — every first-order linear ODE with a stable fixed point traces
# this curve. The terminal value is the fixed point; k is the rate at
# which deviation from the fixed point decays.

Terminal velocity tells you how fast the object eventually falls. The next obvious question is how far has it fallen by time t? — and that’s an integration question. Once $v(t)$ is known, distance is accumulated velocity: $y(t) = ∫_0^t v(s) ds$. The area under the velocity curve, from 0 to $t$, is the distance fallen by then. Plug in $v(s) = v_t · (1 − e^(−k s))$ and the integral is closed-form: $y(t) = v_t · t − (v_t/k)(1 − e^(−kt))$. Two regimes are readable from this: for small $t$, the curve sits below the line $v_t · t$ (you haven’t been falling at terminal speed for the whole interval); for large $t$, the gap stops growing and $y(t)$ is essentially $v_t · t − v_t/k$ — terminal speed times time, minus a constant offset that records “the time you spent accelerating up to terminal.”