University Physics I · Applications of Newton's Laws · 5.6
Drag and Terminal Speed
Drag grows with the speed an object already has, so its acceleration shrinks as it falls. Terminal speed is the speed at which drag has grown enough to balance the weight.
Build the model
Connect the measurement to the mechanism.
Weight and kinetic friction are near enough constant, so they give constant acceleration. Drag is not. It grows with the speed the object already has, so Newton's second law becomes a differential equation in v alone, and the constant-acceleration equations are void from the first instant.
Two models cover most cases: linear drag Fdrag = bv for small, slow objects, and quadratic drag Fdrag = cv² with c = ½ρCdA for everyday objects in air. Setting the acceleration to zero gives the terminal speed — vₜ = mg/b for linear drag, vₜ = √(mg/c) for quadratic drag. Rewriting the second law as a = g[1 − (v/vₜ)ⁿ] then tells the whole story: the acceleration starts at g, shrinks as v climbs, and reaches zero only in the limit.
The same two numbers set the clock, because τ = vₜ/g is the time scale of the approach in both models.
- Simple definition
- Drag is a resistive force from a fluid that grows with speed. Terminal speed is the speed at which drag balances the weight, so the acceleration falls to zero.
- Example
- A 75 kg skydiver with c = 0.42 kg m⁻¹ has drag equal to weight at vₜ = √(735 ÷ 0.42) = 41.8 m s⁻¹, and is within 4% of that speed after about 8.5 s.
Drag from viscous shear. Doubling the speed doubles the force.
b in kg s⁻¹ · η in Pa s, air 1.8 × 10⁻⁵ · use when Re ≲ 1
Drag from shoving fluid aside. Doubling the speed quadruples the force.
c in kg m⁻¹ · Cd is dimensionless · ρ in kg m⁻³
A differential equation in v alone, not a constant-acceleration problem.
Down positive · drag always opposes the velocity
Set a = 0. Linear vₜ scales as m; quadratic vₜ scales as √m.
Units: √(N ÷ kg m⁻¹) = √(m² s⁻²) = m s⁻¹
One expression for both models, and the whole shape of the graph.
a = g at v = 0 · a = 0 at v = vₜ · a < 0 above vₜ
Exponential for linear drag, tanh for quadratic. Same time scale τ.
From rest · τ in s · τ = m/b for linear drag
Drag grows with speed, so acceleration cannot be constant
Weight is constant near the ground and kinetic friction barely changes with speed, so both give constant acceleration. Drag does not. A fluid resists an object more strongly the faster it moves through it, so the resistive force is a function of v. Write the second law along the downward axis and you get m dv/dt = mg − Fdrag(v), which is a differential equation in v alone. From the first instant, every constant-acceleration equation is wrong. The acceleration is largest at release, when the object is slow and drag is small, and it shrinks as the object speeds up.
Two models, and the flow decides which one applies
Linear drag, Fdrag = bv, comes from viscous shear in the layer of fluid dragged along by the surface. For a sphere, Stokes' law fixes the constant: b = 6πηr, with η the dynamic viscosity in Pa s. Quadratic drag, Fdrag = ½ρCdAv², comes from accelerating fluid out of the way and leaving a turbulent wake, so it depends on the fluid density ρ, the frontal area A, and a dimensionless drag coefficient Cd near 0.5 for a smooth sphere and near 1 for a person falling flat. The Reynolds number Re = ρvL/η, with L the diameter for a sphere, picks the regime: below about 1 the flow is viscous and linear drag holds; by a few hundred the wake dominates and quadratic drag holds; between them neither model is clean.
Terminal speed is where drag has grown to equal the weight
Set the acceleration to zero and solve for v. Linear drag gives bvₜ = mg, so vₜ = mg/b. Quadratic drag gives cvₜ² = mg, so vₜ = √(mg/c). The mass dependence differs: at fixed size, doubling the mass doubles vₜ under linear drag but multiplies it by only √2 under quadratic drag. For a 75 kg skydiver flat to the wind, with ρ = 1.2 kg m⁻³, Cd = 1.0 and A = 0.70 m², c = ½(1.2)(1.0)(0.70) = 0.42 kg m⁻¹, so vₜ = √(735 ÷ 0.42) = 41.8 m s⁻¹. A 1 mm-radius raindrop, mass 4.19 × 10⁻⁶ kg with A = 3.14 × 10⁻⁶ m² and Cd = 0.5, gives c = 9.42 × 10⁻⁷ kg m⁻¹ and vₜ = 6.6 m s⁻¹, which is what raindrops of that size actually do — and its Re ≈ 900 confirms the quadratic choice.
Write the acceleration in terms of v/vₜ
Divide the equation of motion by m and factor out g. Linear drag becomes a = g(1 − v/vₜ); quadratic drag becomes a = g[1 − (v/vₜ)²]. The single expression a = g[1 − (v/vₜ)ⁿ] carries both. Read it without solving anything. At v = 0 the acceleration is g. While v < vₜ the bracket is positive, so the object keeps speeding up and can never cross vₜ. At v = vₜ the acceleration is exactly zero. Throw the object downward faster than vₜ and the bracket turns negative, so it decelerates onto the same asymptote from above. The v–t graph therefore rises with a steadily flattening slope, or falls onto the same horizontal line, depending on where it starts.
One time scale runs both approaches: τ = vₜ/g
Solving the two equations for release from rest gives v = vₜ(1 − e(−t/τ)) for linear drag and v = vₜ tanh(t/τ) for quadratic drag, with τ = vₜ/g in both. That τ has a physical reading: the initial slope of the v–t graph is g, so τ is the time the object would need to reach vₜ if drag never came on. The real curve bends away from that tangent immediately. Linear drag reaches 63% of vₜ at t = τ and 95% at 3τ. Quadratic drag is faster, at 76% at τ and 96% at 2τ. For the skydiver, τ = 41.8 ÷ 9.8 = 4.27 s, so the speed is 31.9 m s⁻¹ after one τ and 40.3 m s⁻¹ after 8.5 s. For the raindrop, τ = 0.67 s — it is at terminal speed almost as soon as it forms.
Test which model the data prefers, and know where both stop
Differentiate a measured v–t trace to get a, then plot a against v and against v². Linear drag gives a straight line on the a–v plot; quadratic drag gives a straight line on the a–v² plot. Both lines must extrapolate to a = g at v = 0, and the horizontal intercept is vₜ on the first plot but vₜ² on the second. Judge by the residuals, not by eye. Using the wrong model is not a small error: Stokes' law applied to that 1 mm raindrop, with b = 6πηr = 3.39 × 10⁻⁷ kg s⁻¹, predicts vₜ = 121 m s⁻¹ instead of 6.6. Both forms also assume a fixed shape and a constant coefficient. Cd drifts with Re and drops sharply near Re ≈ 3 × 10⁵ for a smooth sphere, ρ falls with altitude, and buoyancy is ignored — fair for a body in air, where ρair/ρbody ≈ 0.001, but not for a droplet in water.
Change one variable at a time
Make the relationship visible.
Raise the drag constant and watch vₜ fall, the slope-g tangent meet the ceiling sooner, and the acceleration readout sink towards zero as you drag the time marker out.
TERMINAL SPEED vₜ41.8 m s⁻¹
TIME SCALE τ = vₜ/g4.27 s
SPEED AT t (QUADRATIC)30.7 m s⁻¹
ACCELERATION a = g[1−(v/vₜ)²]4.52 m s⁻²
Live interpretationTERMINAL SPEED vₜ: 41.8 m s⁻¹. TIME SCALE τ = vₜ/g: 4.27 s. SPEED AT t (QUADRATIC): 30.7 m s⁻¹. ACCELERATION a = g[1−(v/vₜ)²]: 4.52 m s⁻²
Catch the common trap
Explain before calculating.
A sphere falls through air, and the drag force on it is proportional to v². At the instant its speed is half its terminal speed, what is its acceleration? Take g = 9.8 m s⁻².
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA 0.60 kg basketball falling through air has quadratic drag with c = 0.0120 kg m⁻¹. Find its terminal speed and the time scale τ of the approach. Take g = 9.8 m s⁻².
- At terminal speed the acceleration is zero, so drag has grown to equal the weight: cvₜ² = mg.
- vₜ = √(mg/c) = √(0.60 × 9.8 ÷ 0.0120) = √490 = 22 m s⁻¹.
- τ = vₜ/g = 22.1 ÷ 9.8 = 2.3 s — the time the ball would need to reach vₜ if drag never came on.
- The real curve is v = vₜ tanh(t/τ), so after 5.0 s it is at tanh(2.21) = 0.98 of vₜ, about 22 m s⁻¹ already.
Answervₜ = 22 m s⁻¹; τ = 2.3 s
MediumA 75 kg skydiver has c = 0.42 kg m⁻¹ in a flat, spread posture. At the instant her speed is 25 m s⁻¹, find the drag force and her acceleration, then check the answer against a = g[1 − (v/vₜ)²].
- Weight: mg = 75 × 9.8 = 735 N, down.
- Drag: F(drag) = cv² = 0.42 × 25² = 0.42 × 625 = 262.5 N, up — already 36% of the weight.
- Second law down the axis: a = (735 − 262.5)/75 = 472.5/75 = 6.3 m s⁻², still downward.
- Check: vₜ = √(mg/c) = √(735 ÷ 0.42) = 41.8 m s⁻¹, so (v/vₜ)² = (25/41.8)² = 0.357 and a = 9.8 × (1 − 0.357) = 6.3 m s⁻² — the same number without ever computing the force.
AnswerF(drag) = 2.6 × 10² N; a = 6.3 m s⁻², downward
HardA hailstone falling from rest under quadratic drag has a terminal speed of 12 m s⁻¹. How long does it take to reach 90% of vₜ, what is its acceleration at that instant, and how long would linear drag with the same vₜ take?
- τ = vₜ/g = 12 ÷ 9.8 = 1.22 s.
- Quadratic drag from rest: v = vₜ tanh(t/τ), so tanh(t/τ) = 0.90.
- Invert: t/τ = artanh(0.90) = ½ ln(1.90/0.10) = ½ ln 19 = 1.47, so t = 1.47 × 1.22 = 1.8 s.
- Acceleration there: a = g[1 − (v/vₜ)²] = 9.8 × (1 − 0.90²) = 9.8 × 0.19 = 1.9 m s⁻² — under a fifth of g, though the speed is still climbing.
- Linear drag with the same vₜ: 1 − e(−t/τ) = 0.90 gives t = τ ln 10 = 1.22 × 2.30 = 2.8 s, so the tanh curve gets there first.
Answert = 1.8 s; a = 1.9 m s⁻²; linear drag would need 2.8 s