Apparent Speed of Falling Ball Observed by an Underwater Observer

To find the apparent speed of the falling ball as observed by the fish, we first determine the actual speed of the ball at the given instant.

Step 1: Calculate the actual speed of the ball
The ball is dropped from an initial height of $h_0 = 20\text{ m}$ above the water surface.
At the given instant, the ball is at a height of $h = 12.8\text{ m}$ above the surface.

The downward distance covered by the ball is:
$s = h_0 - h = 20\text{ m} - 12.8\text{ m} = 7.2\text{ m}$

Using the kinematic equation for free fall from rest ($u = 0$):
$v^2 = u^2 + 2gs$
$v^2 = 0 + 2(10\text{ m/s}^2)(7.2\text{ m}) = 144\text{ m}^2/\text{s}^2$
$v = \sqrt{144} = 12\text{ m/s}$

Thus, the actual downward speed of the ball is $v = 12\text{ m/s}$.

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Step 2: Calculate the apparent speed seen by the fish
Let the distance of the ball above the water surface at any instant be $y$.
Since the light rays travel from the rarer medium (air, $\mu_{\text{air}} = 1$) to the denser medium (water, $\mu_{\text{water}} = \mu = \frac{4}{3}$), the apparent height $y'$ of the ball as seen by an underwater observer is given by:
$y' = \mu y$

Differentiating both sides with respect to time $t$:
$\frac{dy'}{dt} = \mu \frac{dy}{dt}$

Taking the magnitude, the apparent speed $v'$ is:
$v' = \mu v$

Substituting the values $\mu = \frac{4}{3}$ and $v = 12\text{ m/s}$:
$v' = \frac{4}{3} \times 12\text{ m/s} = 16\text{ m/s}$

Therefore, the fish observes the ball falling at a speed of $16\text{ m/s}$.

Correct Option: C