When a ball is dropped into a lake from a height 4 . 9m above the...

Using $s = ut + \dfrac{1}{2}a{t}^{2}$

Initially, $u = 0$, so time taken to reach from $A$ to $B$ is $t = \sqrt{\dfrac{2s}{g}} = \sqrt{\dfrac{2 \times 4 . 9}{9 . 8}} = 1s$

Now, using ${v}^{2} = {u}^{2} + 2as$, we have

$v = \sqrt{2gh} = \sqrt{2 \times 9 . 8 \times 4 . 9} = 9 . 8m{s}^{- 1}$

$∵ {t}_{2} = \left(4 - 1\right)s = 3s$

$∴ d = {t}_{2} \times v = 3 \times 9 . 8 = 29 . 4m$