A particle falls in a medium whose resistance is…

A particle falls in a medium whose resistance is proportional to the square of the particle's velocity. Show that the equation of motion is \[ \frac{dv}{dt} = g - kv^2, \] where \(k\) is a constant and \(g\) is the acceleration due to gravity. Integrate this equation and show that \(v\to\sqrt{g/k}\) as \(t\to\infty\).

1. Equation of motion

With \(v\) measured positive downward, gravity contributes \(+mg\) and drag \(-kv^2\) (opposing motion), so Newton's second law \(m\,dv/dt=mg-kv^2\) gives, after dividing by \(m\) and absorbing \(k/m\) into \(k\), \[ \frac{dv}{dt} = g-kv^2. \]

2. Verify the antiderivative

Sympy's \(\texttt{integrate}(1/(g-kv^2),v)\) returns \(\dfrac{1}{2\sqrt{gk}}\Big[\ln(\sqrt{k}v+\sqrt g)-\ln(\sqrt k v-\sqrt g)\Big]\), which is algebraically the same antiderivative the source obtains via partial fractions with \(v_{\text{term}}=\sqrt{g/k}\); the source's partial-fraction step is confirmed correct.

3. Verify the closed-form solution by direct substitution

Substituting \(v(t)=\sqrt{g/k}\tanh\!\big(t\sqrt{gk}\big)\) directly into the original nonlinear equation, Sympy simplifies \[ \frac{dv}{dt}-\big(g-kv^2\big) \] to exactly \(0\) for all \(t\) (using \(1-\tanh^2=\operatorname{sech}^2\) internally) — a genuine check against the ODE itself, not merely a re-derivation. Sympy also confirms \(v(0)=\sqrt{g/k}\tanh(0)=0\), matching the "starts from rest" initial condition.

4. Limiting (terminal) velocity

Sympy's \(\texttt{limit}(v(t),\,t,\,\infty)\) returns \(\sqrt{g/k}\) directly (since \(\tanh(t\sqrt{gk})\to1\)), confirming \[ \lim_{t\to\infty}v(t)=\sqrt{\frac gk}, \] consistent with setting \(dv/dt=0\) in the original equation.