Question

A small spherical ball (obeying Stoke's law for viscous force) is thrown up vertically with a speed $20 \, m \, s^{- 1}$ and is received back by the thrower at the point of projection with a speed $10 \, m \, s^{- 1}$ . Neglecting the buoyant force on the ball, assuming the speed of the ball during its flight to be never equal to its terminal speed and taking the acceleration due to gravity $g=10 \, m \, s^{- 2}$ , find the time of flight of the ball in seconds.

Answer 1

$m\overset{ \rightarrow }{g}+k\overset{ \rightarrow }{v}=\frac{md \overset{ \rightarrow }{v}}{dt}$
$\overset{ \rightarrow }{g} \, dt+\frac{k}{m} \, \overset{ \rightarrow }{v} \, dt=d \, \overset{ \rightarrow }{v}$
Integrating both sides $\displaystyle \int _{0}^{T}\overset{ \rightarrow }{g}dt+\frac{k}{m}\displaystyle \int _{0}^{T}\overset{ \rightarrow }{v}dt=\displaystyle \int _{\overset{ \rightarrow }{u}}^{\overset{ \rightarrow }{v}}d\overset{ \rightarrow }{v}$
$\overset{ \rightarrow }{g}T+0=\overset{ \rightarrow }{v}-\overset{ \rightarrow }{u}$
$gT=v-\left(- u\right)=v+u$
$\Rightarrow T=\frac{v + u}{g}=3s$