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.
(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} 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
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
\[
\lim_{t\to\infty}v(t)=v_{*}. \qquad\blacksquare \]
\[
mv'(t)=mg-kv^2(t),
\] 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} 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
\[
\lim_{t\to\infty}v(t)=v_{*}. \qquad\blacksquare \]
