Two balls are dropped from the same height at  1 second interval of time. The separation between the two balls after 3 seconds of the drop of the 1^st ball is…….$\left(g=9.8m/s^2\right)$

For the motion of the 1^st ball
 $u=0;a=g;t_1=3s$
$\therefore$ The distance travelled by the 1^st ball in 3sec  
$s_1=\frac{1}{2}g{t}_1^2$ 
$=\frac{1}{2}\left(g\right)\left(9\right)=\frac{9}{2}g\mathrm{...........}\left(i\right)$ 
For the motion of 2 ball 
$u=0;a=g;t_2=3-1=2s$ 
The distance travelled by the 2ball in 2s 
$s_2=\frac{1}{2}g{\left(t_2\right)}^2$ 
$=\frac{1}{2}g{\left(2\right)}^2=2g\mathrm{.........}\left(2\right)$ 
Separation between the two balls  $=s_1-s_2$
$\begin{array}{l}=\frac{9}{2}g-2g\\ =\frac{5}{2}g\\ =\frac{5}{2}\times 9.8\\ =24.5\mathrm{m}\end{array}$