1. Find the real speed of the ball at the given instant

The ball is dropped from rest from height $20\,\text{m}$.

At the instant considered, it is $12.8\,\text{m}$ above the water surface. So it has fallen through

$s = 20 - 12.8 = 7.2\,\text{m}$

Using

$v^2 = u^2 + 2gs$

with $u=0$, $g=10\,\text{m/s}^2$:

$v^2 = 2(10)(7.2) = 144$

$v = 12\,\text{m/s}$

So the actual speed of the ball is $12\,\text{m/s}$.

2. Find the apparent height of the ball as seen by the fish

The fish is inside water and the ball is in air. For an object in air seen from water across a plane surface, the apparent distance from the surface is magnified by the refractive index of water:

$h' = \mu h$

where $\mu = \frac{4}{3}$ and $h = 12.8\,\text{m}$.

Thus,

$h' = \frac{4}{3}\times 12.8$

$h' = 17.066\ldots\,\text{m}$

3. Relate apparent speed to real speed

Since

$h' = \mu h$

Differentiating with respect to time,

$v' = \mu v$

So the speed seen by the fish is

$v' = \frac{4}{3}\times 12 = 16\,\text{m/s}$

4. Check options

• A: $9\,\text{m/s}$ ❌
• B: $12\,\text{m/s}$ ❌
• C: $16\,\text{m/s}$ ✅
• D: $21.33\,\text{m/s}$ ❌

Hence, the correct option is C.