Answer
$\frac{mg}{6πηR}$
Work Step by Step
There are two forces acting on the particle of mass $m$:
$1.$ The gravitational force $(mg)$ directed vertically downward,
$2.$ and the drag force $(6πηRs)$ directed vertically upward.
Now, according to Newton’s second law of motion,
$m\frac{ds}{dt}=mg-6πηRs$
or, $\frac{ds}{dt}=g-\frac{6πηRs}{m}$ ..............$(1)$
Let, $\gamma=\frac{6πηR}{m}$. Then eq, $1$ becomes
$\frac{ds}{dt}=g-\gamma s $
or, $\frac{ds}{dt}=-\gamma(s-\frac{g}{\gamma})$
or, $\frac{ds}{(s-\frac{g}{\gamma})}=-\gamma dt$
Integrating both side, we get
$\int_{0}^{s(t)}\frac{ds}{(s-\frac{g}{\gamma})}=-\gamma\int_{0}^{t}dt$
or, $\ln \bigg({\frac{s(t)-\frac{g}{\gamma}}{-\frac{g}{\gamma}}}\bigg)=-\gamma t$
or, ${\frac{s(t)-\frac{g}{\gamma}}{-\frac{g}{\gamma}}}=e^{-\gamma t}$
or, $s(t)=\frac{g}{\gamma}(1-e^{-\gamma t})$
or, $s(t)=\frac{mg}{6πηR}(1-e^{-\frac{6πηR}{m} t})$
For large values of $t$ $(t\to \infty )$ , the term $e^{\frac{6πηR}{m} t}\to0$, and the particle reaches a terminal velocity: $s_\infty=\frac{mg}{6πηR}$
Therefore, the terminal velocity of the particle is given by $\frac{mg}{6πηR}$
