A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2m while applying the force and the ball goes up to 2 m height further, find the magnitude of the force in newton. (Use  .$\mathrm{g}=10{\mathrm{ms}}^{-2})$

For the motion of ball, just after the throwing 
$\mathrm{v}=0,\mathrm{s}=2\mathrm{m},\mathrm{a}=-\mathrm{g}=-10{\mathrm{ms}}^2$
${\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{as}$For upward journey
$\Rightarrow -{\mathrm{u}}^2=2(-10)\times 2\Rightarrow {\mathrm{u}}^2=40$
When the ball is in the hands of the thrower
$\begin{array}{l}\mathrm{u}=0,\mathrm{v}=\sqrt{40}{\mathrm{m}}^{-1}\\ \mathrm{s}=0.2\mathrm{m}\\ {\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{as}\\ \Rightarrow 40-0=2\left(\mathrm{a}\right)0.2\Rightarrow \mathrm{a}=100\mathrm{m}/{\mathrm{s}}^2\\ \therefore \mathrm{F}=\mathrm{ma}=0.2\times 100=20\mathrm{N}\\ \Rightarrow \mathrm{N}-\mathrm{mg}=20\Rightarrow \mathrm{N}=20+2=22\mathrm{N}\end{array}$