Linear Drag and Terminal Velocity suits a bead in syrup. A skydiver, a baseball or a car pushes air aside, and the drag then grows with the square of the speed. You can explore its effect on launches in the Trebuchet lab, where drag pulls the best angle below 45°.
Key idea
At everyday speeds in air, $$F_D = \tfrac{1}{2}C\rho Av^2,$$ where $\rho$ is the air's density (about $1.2\ \text{kg/m}^3$ near sea level), $A$ the cross-sectional area facing the flow, and $C$ the drag coefficient (about 0.3 to 0.5 for a ball, about 1 for a spread-out skydiver). Setting $F_D = mg$ gives the terminal speed $$v_T = \sqrt{\dfrac{2mg}{C\rho A}}.$$
Why it works
Each second, a moving object sweeps through a column of air of volume $Av$ and mass $\rho Av$, and shoves it to a speed comparable to $v$. The momentum given to the air each second, which by Newton's third law is the force back on the object, is about $(\rho Av)\cdot v = \rho Av^2$. The factor $\tfrac{1}{2}C$ captures how the shape steers the air. So doubling the speed quadruples the drag.
Worked example: A skydiver
An $80\ \text{kg}$ skydiver spread out flat has $C = 1.0$ and $A = 0.70\ \text{m}^2$. Find her terminal speed in air with $\rho = 1.2\ \text{kg/m}^3$.
- $$v_T = \sqrt{\dfrac{2(80)(9.8)}{(1.0)(1.2)(0.70)}} = \sqrt{\dfrac{1568}{0.84}} = \sqrt{1867}$$
- $v_T = 43.2\ \text{m/s}$, about 155 km/h. Head-first, with a smaller $A$ and $C$, she falls much faster.
Answer: $43.2\ \text{m/s}$.
Worked example: Heavier but the same shape
A steel ball and a wooden ball have the same size, but the steel one is 9 times as massive. Compare their terminal speeds in air.
- With $C$, $\rho$ and $A$ equal, $v_T \propto \sqrt{m}$.
- $\sqrt{9} = 3$: the steel ball's terminal speed is 3 times as large, so in air it pulls ahead.
Answer: The heavier ball's terminal speed is 3 times as big.
Common mistake
Concluding from experience in air that heavier objects fall faster in a vacuum too. Without air there is no drag, and gravity's pull $mg$ divided by the inertia $m$ gives the same $a = g$ for every mass, as astronauts showed with a hammer and a feather on the Moon. Check: in the formula for $v_T$, the mass matters only because drag exists; with $\rho = 0$, $v_T$ is infinite for everyone and all fall together.
Solving $m\,dv/dt = mg - \tfrac{1}{2}C\rho Av^2$ by separation (with partial fractions) gives $v(t) = v_T\tanh(gt/v_T)$, the curve in the picture. Unlike linear drag, the approach to $v_T$ is still exponential in the end, but the early fall looks almost like free fall, because drag is tiny at low speed.