Question

A freely falling body acquires a velocity $vm{s}^{-1}$ in falling through a distance of $80m$. How much further distance should it fall, so as to acquire a velocity of $2vm{s}^{-1}$? (Take $g=10m{s}^{-2}$)

Answer 1

$v^2=2gs$

$v=\sqrt{2gs}$

$s=80,V=v$

$V=\sqrt{2\times 10\times 80}=\sqrt{1600}=40m/s$

$2V=\sqrt{2g{S}^{\prime }}$

$24\times 1600=2\times 10\times S^{\prime }$

$S^{\prime }=320m$

Further distance =$320-40=280m$