Problem — Fall with air resistance

Problem

A body of mass $m$ falls with air resistance proportional to the velocity: $mv' = mg - kv$. Solve for $v(t)$ with $v(0)=0$ and find the terminal velocity.

Solution

We rewrite it as $v' = g - \dfrac{k}{m}v$: it is separable. We separate: $\frac{dv}{g-\frac{k}{m}v} = dt.$ Integrating (the left-hand side gives $-\frac{m}{k}\ln\bigl|g-\frac{k}{m}v\bigr|$) and imposing $v(0)=0$ we obtain: $\boxed{v(t) = \frac{mg}{k}\left(1 - e^{-\frac{k}{m}t}\right)}$ The terminal velocity is found as $t\to\infty$ (or by setting $v'=0$): $\boxed{v_{\lim} = \frac{mg}{k}}$ The velocity grows and approaches the limiting value exponentially, when the resistance $kv$ exactly balances the weight force $mg$.

Links

Topics: Differential equations
Concepts: Decay · First-order linear ODE · Separable variables
Functions: Exponential function
Skills: Integrating · Modelling · Solving equations
Exercise type: Modelling problem