When a ball is dropped into a lake from a height 4 . 9   m above the water level, it hits the water…

Using $s=ut+\frac{1}{2}a{t}^2$

Initially, $u=0$, so time taken to reach from $A$ to $B$ is  $t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2\times 4.9}{9.8}}=1 \mathrm{s}$

Now, using $v^2=u^2+2as$, we have

$v=\sqrt{2\mathrm{gh}}=\sqrt{2\times 9.8\times 4.9}=9.8 \mathrm{m} {\mathrm{s}}^{-1}$ 

$\because t_2=\left(4-1\right) \mathrm{s}=3 \mathrm{s}$  

$\therefore d=t_2\times v=3\times 9.8=29.4 \mathrm{m}$