1. Interpret the condition

The ball falls freely through air from rest over a height $h$, so just before entering water its speed is

$v = \sqrt{2gh}$

After entering water, its velocity does not change and it continues with the same constant velocity.
This means that this speed must be equal to the terminal velocity in water.

2. Forces on the ball in water

Since it moves with constant velocity in water,

$\text{net force} = 0$

For a small sphere in a viscous fluid, viscous drag is given by Stokes' law:

$F_v = 6\pi \eta r v$

Other forces:

• Weight: $W = \frac{4}{3}\pi r^3 \rho g$
• Buoyant force: $B = \frac{4}{3}\pi r^3 \rho_w g$

So at terminal velocity,

$W = B + F_v$

$\frac{4}{3}\pi r^3 (\rho-\rho_w)g = 6\pi \eta r v$

Thus,

$v = \frac{2r^2(\rho-\rho_w)g}{9\eta}$

3. Substitute values

Given:

• Radius: $r = 0.1\text{ mm} = 10^{-4}\text{ m}$
• Density of ball: $\rho = 10^4\text{ kg m}^{-3}$
• Density of water: $\rho_w = 10^3\text{ kg m}^{-3}$
• $g=10\text{ m s}^{-2}$
• $\eta = 1.0\times 10^{-5}\text{ N s m}^{-2}$

Now,

$v = \frac{2(10^{-4})^2(10^4-10^3)(10)}{9(10^{-5})}$

$= \frac{2\times 10^{-8}\times 9\times 10^3\times 10}{9\times 10^{-5}}$

$= \frac{18\times 10^{-4}}{9\times 10^{-5}}$

$= 2\times 10 = 20\text{ m s}^{-1}$

So the speed of the ball just before entering water must be

$v = 20\text{ m s}^{-1}$

4. Relate this to free fall height

From free fall,

$v^2 = 2gh$

$20^2 = 2\cdot 10 \cdot h$

$400 = 20h$

$h = 20\text{ m}$

5. Final answer

$\boxed{20}$

The required height is $20\,$m.