A ball is dropped from a height. Another ball is dropped from the same height after  4s. Their separation after 2 more seconds is ………..  $m \left(g=9.8m/s^2\right)$

First ball travelled for 6s and the second ball for 2s
$\therefore$ Distance travelled by the first body is  $s_1=\frac{1}{2}g{t}_1^2=\frac{1}{2}\times g\times 36=18g$    
Distance travelled by the second body is   $s_2=\frac{1}{2}g{t}_2^2=\frac{1}{2}\times g\times 4=2g$   
$\therefore$ Their separation is $18g-2g$ 
$\begin{array}{l}=16g\\ =16\times 9.8=156.80\mathrm{m}\end{array}$