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Q.
A ball of mass $m$ is thrown upwards with a velocity $v$. If air exerts an average resisting force $F$, the velocity with which the ball returns to the thrower is
Laws of Motion
Solution:
For upward motion,
Net force $=m g+F$
$\therefore $ Retardation $(a)=\frac{m g+F}{m}$
Distance, $s=\frac{v^{2}}{2 a}=\frac{v^{2} m}{2(m g+F)} \,\,\,\,\, ....(i)$
For downward motion, net force $=m g-F$
$\therefore $ Acceleration $\left(a^{\prime}\right)=\frac{m g-F}{m}$
Distance, $s^{\prime}=\frac{v^{\prime 2}}{2 a^{\prime}}=\frac{v^{\prime 2} m}{2(m g-F)} \,\,\,\,\,\, ...(ii)$
As $s=s^{\prime} \therefore \frac{v^{2} m}{2(m g+F)}=\frac{v^{\prime 2} m}{2(m g-F)}$
or $v^{\prime}=v \sqrt{\frac{m g-F}{m g+F}}$