Terminal Velocity: Why Objects Stop Accelerating & Calculations

Terminal velocity is the constant speed reached by a falling object when the drag force equals its weight, giving zero net force and zero acceleration. The drag force on an object moving through a fluid is F_drag = ½ρCdAv², where ρ is fluid density, Cd is the drag coefficient, A is the cross-sectional area, and v is speed. As an object falls faster, drag increases until it exactly balances weight (mg) — at that point, net force is zero and the object continues at constant terminal velocity.

Terminal velocity is why a skydiver doesn't keep accelerating indefinitely, why a raindrop falls at a few metres per second rather than at hundreds, and why a feather floats gently down while a stone plummets. The terminal speed depends critically on the object's size, shape, mass, and the fluid it moves through — a design insight exploited by parachutes, aerodynamic vehicles, and even dandelion seeds.

How Terminal Velocity is Reached

When an object is first released from rest and begins falling:

1. Initially: v = 0, drag = 0. Net force = mg (weight only). Acceleration = g downward.
2. As speed increases: drag force increases (∝ v²). Net force = mg − drag. Acceleration decreases but is still positive — the object still accelerates, just more slowly.
3. At terminal velocity: drag = mg. Net force = 0. Acceleration = 0. Speed is constant.

The velocity-time graph shows a curve that rises steeply at first (acceleration ≈ g) then flattens as drag builds, approaching an asymptote at v_t. The object never mathematically reaches terminal velocity in finite time, but approaches it exponentially.

The Drag Force Formula

Where:

• ρ = density of the fluid (kg/m³) — air: ≈ 1.2 kg/m³, water: 1000 kg/m³
• Cd = drag coefficient (dimensionless) — depends on shape: sphere ≈ 0.47, flat plate ≈ 1.28, streamlined body ≈ 0.04
• A = cross-sectional area perpendicular to motion (m²)
• v = speed relative to the fluid (m/s)

The v² dependence is crucial: doubling speed quadruples drag. This is why the top speed of a car increases much more slowly than its engine power — aerodynamic drag increases so rapidly that enormous power increases produce only modest speed gains.

Terminal Velocity Formula

At terminal velocity, F_drag = mg:

Terminal velocity increases with:

• Greater mass (heavier objects fall faster at terminal) — a lead ball vs a foam ball of the same size
• Smaller cross-section — a diving position vs spread-eagle position for a skydiver
• Lower drag coefficient — streamlined vs blunt shapes
• Lower fluid density — falling through air vs water

4 Worked Examples

Example 1 — Skydiver terminal velocity

Problem: A skydiver of mass 80 kg falls in spread-eagle position with cross-sectional area 0.7 m² and drag coefficient Cd = 1.0. Find the terminal velocity in air (ρ = 1.2 kg/m³).

Solution:
v_t = √(2mg / ρCdA) = √(2 × 80 × 9.81 / (1.2 × 1.0 × 0.7))
= √(1569.6 / 0.84) = √1868.6 = 43.2 m/s ≈ 43 m/s (155 km/h)

Example 2 — Skydiver with parachute

Problem: The same skydiver deploys a parachute with A = 50 m² and Cd = 1.3. Find the new terminal velocity.

Solution:
v_t = √(2 × 80 × 9.81 / (1.2 × 1.3 × 50))
= √(1569.6 / 78) = √20.12 = 4.49 m/s ≈ 4.5 m/s (16 km/h)
The parachute reduces terminal velocity from 43 m/s to 4.5 m/s — a factor of ~10 — making landing survivable.

Example 3 — Raindrop terminal velocity

Problem: A raindrop has radius 1.5 mm (A = π × (1.5 × 10⁻³)² = 7.07 × 10⁻⁶ m²), mass 1.41 × 10⁻⁵ kg, and Cd = 0.47. Find its terminal velocity. (ρ_air = 1.2 kg/m³)

Solution:
v_t = √(2 × 1.41 × 10⁻⁵ × 9.81 / (1.2 × 0.47 × 7.07 × 10⁻⁶))
= √(2.767 × 10⁻⁴ / 3.987 × 10⁻⁶) = √69.4 = 8.33 m/s ≈ 8.3 m/s
Raindrops fall at 7–9 m/s — if terminal velocity weren't so low, rain would be dangerous.

Example 4 — Net force at a given speed

Problem: The skydiver in Example 1 reaches 20 m/s before terminal velocity. Find the net downward force and acceleration at this speed.

Solution:
Weight = mg = 80 × 9.81 = 784.8 N
Drag = ½ρCdAv² = ½ × 1.2 × 1.0 × 0.7 × 400 = 168 N
Net force = 784.8 − 168 = 616.8 N downward
Acceleration = F/m = 616.8/80 = 7.71 m/s² (less than g = 9.81 m/s², because drag is already significant)

Time to Reach Terminal Velocity — The Mathematics

The differential equation of motion for a falling object with drag F = bv (linear drag, valid at low speeds):

Solution: v(t) = v_t(1 − e^{−bt/m}) = v_t(1 − e^{−t/τ})

Where v_t = mg/b is terminal velocity and τ = m/b is the time constant. After time τ, the object reaches 63% of terminal velocity; after 5τ, it reaches 99.3% — essentially terminal velocity.

For the more physical quadratic drag F = ½ρCdAv² (valid at high speeds, like skydivers): v(t) = v_t × tanh(gt/v_t), where tanh is the hyperbolic tangent function. The result is the same qualitative shape — exponential approach to v_t — but with the exact formula differing. At t = v_t/g, the object has reached tanh(1) ≈ 76% of terminal velocity.

Stokes' Law for Small Spheres at Low Speed

For very small spheres at low speed (Re < 1), viscous drag dominates and the drag force is:

Where η is the dynamic viscosity of the fluid and r is the sphere radius. Terminal velocity from F_drag = mg (buoyancy-corrected):

This formula is the basis of the Stokes' method for measuring viscosity: drop a small ball bearing of known size and density through the fluid, measure its terminal velocity, and calculate η. Used for viscous oils, glycerol, and biological fluids. Note that v_t ∝ r² — a ball twice as large falls four times as fast (at the same low Re).

Worked Example 5 — Raindrop terminal velocity

Problem: A spherical raindrop of radius 1.5 mm falls through air at sea level. Using Stokes' law (checking validity afterwards), find the terminal velocity. (ρ_water = 1000 kg/m³, ρ_air = 1.2 kg/m³, η_air = 1.8 × 10⁻⁵ Pa·s)

Solution:
m = (4/3)π(1.5 × 10⁻³)³ × 1000 = 1.414 × 10⁻⁵ kg
v_t = 2r²(ρ_sphere − ρ_fluid)g/(9η) = 2(1.5 × 10⁻³)²(1000 − 1.2)(9.81)/(9 × 1.8 × 10⁻⁵)
= 2(2.25 × 10⁻⁶)(998.8)(9.81)/(1.62 × 10⁻⁴)
= 2 × 2.25 × 10⁻⁶ × 9798.2 / 1.62 × 10⁻⁴ = 4.409 × 10⁻² / 1.62 × 10⁻⁴ = 272 m/s
This is absurd — clearly Stokes' law is not valid here! Check Re = ρ_air × v_t × 2r/η = 1.2 × 272 × 0.003/1.8 × 10⁻⁵ = 54,400 ≫ 1. Stokes' law fails for raindrops — the quadratic drag formula must be used instead (giving ~7–9 m/s as shown in the main article). Stokes' law is valid only for Re < 0.1 (e.g. aerosol particles, bacteria, fine dust).

Reynolds Number — Predicting Flow Regime

The Reynolds number Re determines whether drag is linear (viscous-dominated, Stokes) or quadratic (inertia-dominated):

Where L is a characteristic length (diameter for spheres). Re < 1: viscous drag (Stokes' law applies). 1 < Re < 1000: transition regime. Re > 1000: turbulent/inertial drag (F ∝ v²). Re > ~500,000: turbulent boundary layer (drag crisis — CD drops suddenly as turbulent BL reattaches).

The Reynolds number also determines whether flow is laminar (smooth) or turbulent (chaotic). Laminar in pipe flow for Re < ~2000; turbulent above ~4000; transitional between. Turbulent flow dissipates far more energy — a factor of ~5–10 more pressure drop for the same flow rate at the same Re. This is why blood flow in capillaries (Re ≈ 0.001) is smoothly laminar while flow in the aorta (Re ≈ 4000) is turbulent, causing the characteristic "turbulent bruit" audible with a stethoscope in some conditions.

Worked Example 6 — Comparing terminal velocities

Problem: A lead sphere (ρ = 11,340 kg/m³) and an aluminium sphere (ρ = 2,700 kg/m³) of equal radius 2 cm fall through air. Using F = ½ρ_air CdAv² with Cd = 0.47, find the ratio of their terminal velocities. (ρ_air = 1.2 kg/m³)

Solution:
At terminal velocity: v_t = √(2mg/(ρ_air CdA)) = √(2(4/3)πr³ρ_sphere g/(ρ_air Cd πr²)) = √(8rρ_sphere g/(3ρ_air Cd))
Since r, g, ρ_air, and Cd are the same for both, v_t ∝ √ρ_sphere.
v_lead/v_aluminium = √(ρ_lead/ρ_aluminium) = √(11340/2700) = √4.2 = 2.05
The lead ball falls about twice as fast at terminal velocity. v_lead = √(8×0.02×11340×9.81/(3×1.2×0.47)) = √(3181) = 56.4 m/s
v_aluminium = 56.4/2.05 = 27.5 m/s

Skydiving Physics in Practice

Professional skydivers exploit changes in body orientation to control their fall rate. Spread-eagle position (maximum area A ≈ 0.7 m², maximum Cd ≈ 1.0): terminal velocity ~55 m/s (200 km/h). Head-down position (minimum area A ≈ 0.05 m², streamlined Cd ≈ 0.3): terminal velocity ~150–200 m/s (540–720 km/h). The world record for the fastest skydiving speed (without a special suit) is about 525 km/h by a free-flier in head-down configuration. Felix Baumgartner's stratospheric jump in 2012 reached 1,357 km/h — supersonic — because at 39 km altitude ρ_air is less than 1% of sea-level density, so the drag force at any speed is tiny and terminal velocity is orders of magnitude higher than at ground level.

Exam Summary for Terminal Velocity

Terminal velocity is reached when drag force equals driving force (net force = 0, acceleration = 0). Formula: v_t = √(2mg/(ρ_fluid CdA)) for quadratic drag (F = ½ρCdAv²). Factors that increase v_t: greater mass (heavier objects fall faster at terminal); smaller cross-sectional area; lower drag coefficient (streamlined shape); lower fluid density. Factors that decrease v_t: smaller mass; larger area; higher Cd; higher fluid density (water vs air). The velocity-time graph: starts at gradient g (no drag initially), curve becomes less steep as drag builds, asymptotically approaches v_t (gradient → 0). The key conceptual point: the object reaches constant speed, not zero speed. "Terminal" means constant, not stopped.

Common misconception: heavier objects always reach the ground first when dropped from the same height. In vacuum: true — all objects fall at the same rate regardless of mass (Galileo). In air with terminal velocity: it depends. Two objects of different mass but same shape and size reach different terminal velocities; if the fall height is small (both below terminal velocity throughout), they hit the ground together. If the height is large (one reaches terminal, the other doesn't), the heavier object arrives first. A feather and a hammer dropped on the Moon (no air) land simultaneously — as demonstrated by the Apollo 15 astronaut in 1971.

Terminal velocity is one of the clearest demonstrations of Newton's second law in a non-trivial scenario. At terminal velocity, Newton's first law applies (constant velocity, zero net force) even though two large forces — weight and drag — are both acting. The balance between them is not a static equilibrium (the forces are large and equal) but a dynamic one maintained only at that specific speed. Drop below terminal velocity and drag decreases, net downward force reappears, object accelerates back up. Exceed terminal velocity (e.g. by opening a parachute in a fast fall) and drag exceeds weight, net force is upward, object decelerates. Terminal velocity is a stable equilibrium of the velocity, not the position — a fundamental distinction between static and dynamic equilibria in physics.

In summary, terminal velocity is the speed at which aerodynamic drag exactly balances the net downward force on a falling object. It is reached asymptotically — the object continuously decelerates (i.e. acceleration decreases, not velocity) until velocity stabilises. The formula v_t = √(2mg/ρCdA) shows all the dependencies clearly: double the mass → speed increases by √2; quadruple the area → speed halves. Design applications range from parachutes (large A → low v_t → safe landing) to streamlined vehicles (low Cd → high v_t for sports cars, aircraft) to sports balls (dimples on a golf ball create turbulent boundary layer that reduces CD and increases v_t, allowing longer drives). Terminal velocity physics is applied wherever objects move through fluids — from meteorites to submarines, from pollen to spacecraft re-entry capsules.

The velocity-time graph for a falling object with quadratic drag has a characteristic sigmoid-like shape (actually v = v_t tanh(gt/v_t)): steepest gradient at t = 0 (acceleration = g, no drag yet), decreasing gradient as drag builds, flattening asymptotically to v_t. The corresponding acceleration-time graph starts at g and decreases exponentially to zero. Plotting acceleration vs velocity directly gives a straight line (for linear drag) or a curve (quadratic drag): a = g − bv/m or a = g − kv²/m, both approaching zero as v → v_t. These graphs are exam favourites — always label the initial gradient (= g), the asymptote (= v_t), and the shape (decreasing gradient, asymptotic approach, not sigmoid approaching from below for acceleration).

One of the most important practical applications of terminal velocity physics is in the design of safety systems. Automobile crumple zones extend the collision time; airbags extend the deceleration time further. But terminal velocity physics directly governs the design of parachute systems, emergency aircraft exit slides, elevator safety brakes, and industrial assembly line shock absorbers — anywhere a controlled deceleration is needed. The governing constraint is always the same: the deceleration force on the occupant/package must remain within survivable or acceptable limits. Setting this force equal to a maximum allowable value and applying F_drag = ½ρCdAv² determines the minimum parachute area or the required cushioning depth — terminal velocity physics as life-safety engineering.