Subsection 4.5.2 Motion in Air Drag
When an object of mass \(m\) is thrown vertically down ward with air resistance given as \(f=cv\text{.}\) Equation of motion with air resistance, \(\sum F=ma.\)
\begin{equation}
F=ma=w-f = mg-cv \tag{4.5.6}
\end{equation}
\begin{equation*}
a=g-\frac{c}{m}v
\end{equation*}
\begin{equation}
a= g-kv \qquad \text{[where $k=\frac{c}{m}$]} \tag{4.5.7}
\end{equation}
\begin{equation*}
\text{or,}\quad \frac{\,dv}{\,dt} = g-kv
\end{equation*}
\begin{equation*}
\text{or,}\quad \int \frac{\,dv}{g-kv} =\int \,dt
\end{equation*}
\begin{equation*}
\text{or,}\quad \frac{\ln(g-kv)}{-k} =t + A
\end{equation*}
\begin{equation*}
\text{or,}\quad \ln(g-kv) =-kt-kA
\end{equation*}
\begin{equation}
\text{or,} \quad g-kv = e^{-kt}\cdot e^{-kA} \tag{4.5.8}
\end{equation}
At \(t=0, \quad v=u,\) initial velocity with which the ball is thrown.
\begin{equation}
\therefore g-ku = e^{-kA} \tag{4.5.9}
\end{equation}
Now from eqn. (4.5.8), we have -
\begin{equation*}
g-kv = e^{-kt} (g-ku)
\end{equation*}
\begin{equation*}
-kv = -g+e^{-kt} (g-ku)
\end{equation*}
\begin{equation}
\therefore v(t) = \frac{g}{k}-\left(\frac{g}{k}-u\right)e^{-kt} \tag{4.5.10}
\end{equation}
Again after a very long time, \(t \to \infty\text{,}\) \(e^{-kt} \to e^{-\infty}=0\text{.}\)
\begin{equation*}
\therefore v(t) = \frac{g}{k} =\frac{gm}{c} =constant = v_{t}
\end{equation*}
where \(v_{t} = \frac{gm}{c}\) is called the terminal velocity.
\begin{equation}
\therefore v(t) = v_{t}+\left(u-v_{t}\right)e^{-kt} \tag{4.5.11}
\end{equation}
and eqn. (4.5.7) becomes
\begin{equation}
a(t) = \frac{c}{m}\left(v_{t}-v\right) \tag{4.5.12}
\end{equation}
Now at \(t=0,\) eqn.(4.5.11) gives
\begin{equation*}
v(t) =v_{t}+\left(u-v_{t}\right)\cdot 1
\end{equation*}
\begin{equation*}
\therefore v(t) =u
\end{equation*}
At \(t \to \infty\) eqn. (4.5.10) gives -
\begin{equation*}
v(t) = v_{t}+\left(u-v_{t}\right)\cdot 0
\end{equation*}
\begin{equation*}
\therefore v(t) = v_{t}
\end{equation*}
Again from eqn. (4.5.11),
\begin{equation*}
v(t) =\frac{\,dy(t)}{\,dt}= v_{t}+\left(u-v_{t}\right)e^{-kt}
\end{equation*}
\begin{equation*}
y(t) =\int\,dy=\int v_{t}\,dt+\int\left(u-v_{t}\right)e^{-kt}\,dt
\end{equation*}
\begin{equation}
\therefore y(t) = v_{t}t -\frac{1}{k}\left(u-v_{t}\right)e^{-kt} +B \tag{4.5.13}
\end{equation}
At \(t=0, \quad y=y_{o}\) (initial position) and \(v=u\) (initial velocity).
\begin{equation*}
\therefore y_{o} = -\frac{1}{k}\left(u-v_{t}\right)\cdot 1+B
\end{equation*}
\begin{equation*}
\therefore B = y_{o} -\frac{1}{k}\left(v_{t}-u\right)
\end{equation*}
Therefore from eqn. (4.5.13), we have -
\begin{equation*}
y=y_{o}-\frac{1}{k}\left(v_{t}-u\right)+v_{t}\cdot t - \frac{1}{k}\left(v_{t}-u\right)e^{-kt}
\end{equation*}
\begin{equation*}
y= y_{o}+v_{t}\cdot t-\frac{1}{k}\left(v_{t}-u\right)\{1-e^{-kt}\}
\end{equation*}
\begin{equation}
\therefore y= y_{o}+v_{t}\cdot t+\frac{1}{k}\left(v_{t}-u\right)\{e^{-kt}-1\} \tag{4.5.14}
\end{equation}
Without air resistance, the position of an object is given by
\begin{equation*}
y=y_{o}+ut+\frac{1}{2}gt^{2}
\end{equation*}
If \(y_{o}=0, \quad y=ut+\frac{1}{2}gt^{2}\) for falling object. In case if the object is thrown upward, then
\begin{equation*}
y= y_{o}-v_{ter}\cdot t+\frac{1}{k}\left(u+v_{t}\right)\{1-e^{-kt}\}
\end{equation*}
If \(y_{o}=0,\quad y =-v_{t}\cdot t + \frac{1}{k}\left(u+v_{t}\right)\{1-e^{-kt}\}. \) Without air resistance in upward motion, \(y=ut-\frac{1}{2}gt^{2}\text{.}\) [since \(c=0=k\)]
