Free Falling 2

Problem 2.

An object (its mass is $m$) falls through the air toward Earth. Assuming that the only forces acting on the object are gravity ($g$ is the gravity constant) and air resistance (proportional to the square of the speed of the object).

(a) Determine the speed $v(t)$ of the object.
(b) Show that $\displaystyle{\lim_{t\to\infty}}v(t)=v_{*}$ independently of the initial speed $v_0$, where $v_{*}$ is the equilibrium speed.

Show Solution

(a)  Applying Newton's second law we obtain the differential equation
$$mv'(t)=mg-kv^2(t),$$ where $m,g,k>0$ are constants and $v(0)=v_0\gt v_{*}$. (see below).
The equilibrium speed $v_{*}$ is
$$v_{*}=\sqrt\frac{mg}{k}.$$ Separating the variables in the differential equation we have
$$\begin{equation} \label{eq:diffeq} \frac{m}{kv{_*}^2-kv^2}\,dv=1\,dt. \end{equation}$$

From partial fraction decomposition

it follows $$\frac{m}{kv{_*}^2-kv^2}=\frac{m}{2kv_{*}(v+v_{*})}-\frac{m}{2kv_{*}(v-v_{*})}.$$ Integrating in $\eqref{eq:diffeq}$ both sides we obtain

$$\begin{align*} &\Step{1A}{\frac{m}{2kv_{*}}\ln\,\left| \frac{v+v_{*}}{v-v_{*}} \right| =t+c,}\\ &\Step{2A}{\ln\,\left| \frac{v+v_{*}}{v-v_{*}}\right|=\frac{2kv_{*}t}{m}+c,} \\ &\Step{3A}{\frac{v+v_{*}}{v-v_{*}}=ce^{2kv_{*}t/m}.} \\ \end{align*}$$

Solving for $v$ $$v(t)=v_{*}\frac{1+ce^{2kv_{*}t/m}}{-1+ce^{2kv_{*}t/m}}.$$ From the initial condition we find $$c=\frac{v_{*}+v_0}{-v_{*}+v_0}.$$ Substituting this value of $c$ we obtain
$$\begin{equation} \label{eq:sol} v(t)=v_{*}\frac{(v_{*}+v_0)e^{2kv_{*}t/m}-v_{*}+v_0}{(v_{*}+v_0)e^{2kv_{*}t/m}+v_{*}-v_0}. \end{equation}$$ (b)  From $\eqref{eq:sol}$ it follows
$$\lim_{t\to\infty}v(t)=v_{*}. \qquad\blacksquare$$