A ball is dropped from a height of 20 ,m abov | vedclass

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(C) Let $h$ be the height of the ball above the water surface. The velocity of the ball at height $h$ is given by $v_b = \sqrt{2g(H - h)}$, where $H =

Vedclass · Accepted Answer

$16$

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(C) Let $h$ be the height of the ball above the water surface. The velocity of the ball at height $h$ is given by $v_b = \sqrt{2g(H - h)}$, where $H = 20 \,m$ is the initial height and $h = 12.8 \,m$.$v_b = \sqrt{2 	imes 10 	imes (20 - 12.8)} = \sqrt{20 	imes 7.2} = \sqrt{144} = 12 \,m/s$.When an object is in a rarer medium (air) and viewed from a denser medium (water), the apparent height $h'$ is given by $h' = \mu h$, where $\mu = 4/3$ is the refractive index of water.Therefore, the apparent velocity $v_a$ as seen by the fish is $v_a = \frac{d}{dt}(h') = \mu \frac{dh}{dt} = \mu v_b$.$v_a = (4/3) 	imes 12 \,m/s = 16 \,m/s$.

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