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(C) Let the height of the ball above the water surface be $x$. The apparent height $x'$ of the ball as seen by the fish is given by $x' = \mu x$, wher

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$16$

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(C) Let the height of the ball above the water surface be $x$. The apparent height $x'$ of the ball as seen by the fish is given by $x' = \mu x$, where $\mu = 4/3$ is the refractive index of water.The speed of the ball as seen by the fish is $v' = \frac{dx'}{dt} = \mu \frac{dx}{dt} = \mu v$, where $v$ is the actual speed of the ball.First, we calculate the actual speed $v$ of the ball when it is at a height $x = 12.8 \ m$. The ball falls from $H = 20 \ m$. The distance fallen is $h = 20 - 12.8 = 7.2 \ m$.Using the equation of motion $v^2 = u^2 + 2gh$, with $u = 0$:$v^2 = 2 	imes 10 	imes 7.2 = 144$$v = 12 \ m/s$.Now, the apparent speed $v'$ seen by the fish is:$v' = \mu v = (4/3) 	imes 12 = 16 \ m/s$.

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