A small spherical ball of radius 0 . 1mm and density 10 ^ 4 kg m ^...

Using equation of motion, ${v}^{2} = {u}^{2} + 2as$

${v}^{2} = 0 + 2\left(- g\right)\left(- h\right)$

$⇒ {v}^{2} = 2gh$

Initial velocity of ball is $v = \sqrt{2gh} = \sqrt{20h}$.

If after entering the water this velocity does not change, then this value should be equal to the terminal velocity.

Then, $\sqrt{2gh} = \dfrac{2}{9}\dfrac{{r}^{2}\left(ρ - \sigma\right)g}{η}$, here, $ρ$ is density of ball, $\sigma$ is density of water and $η$ is viscosity.

$⇒ \sqrt{20h} = \dfrac{2}{9}\dfrac{{\left(0 . 1 \times {10}^{- 3}\right)}^{2}\left({10}^{4} - {10}^{3}\right)10}{1 \times {10}^{- 5}} ⇒ \sqrt{20h} = \dfrac{2}{9}\dfrac{{10}^{- 8}\left(10 - 1\right){10}^{3} \times 10}{1 \times {10}^{- 5}}$

$⇒ 20h = 400$

$∴ h = 20m$