Free Falling 1

Problem 1.

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 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
$$mv'(t)=mg-kv(t),$$ where $m,g,k>0$ are constants and $v(0)=v_0\geq 0$. The equilibrium speed $v_{*}$ is $$v_{*}=\frac{mg}{k}.$$ Separating the variables in the differential equation we have $$\begin{equation} \label{eq:diffeqlin} \frac{m}{kv{_*}-kv}\,dv=1\,dt. \end{equation}$$ Integrating in $\eqref{eq:diffeqlin}$ both sides we obtain

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

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