A fish rising vertically at the rate of 3 , m | vedclass

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(D) Let the velocity of the fish be $v_f = 3 \, m/s$ (upwards) and the actual velocity of the bird be $v_b$ (downwards).Let the depth of the fish be $

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$4.5$

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(D) Let the velocity of the fish be $v_f = 3 \, m/s$ (upwards) and the actual velocity of the bird be $v_b$ (downwards).Let the depth of the fish be $x$ and the height of the bird be $y$ from the water surface.The apparent height of the bird as seen by the fish is $y' = \mu y$,where $\mu = 4/3$.The total apparent distance $h$ of the bird from the fish is $h = x + \mu y$.Differentiating with respect to time $t$,we get the apparent velocity of the bird relative to the fish: $\frac{dh}{dt} = \frac{dx}{dt} + \mu \frac{dy}{dt}$.Here,$\frac{dh}{dt} = 9 \, m/s$ (the rate at which the fish sees the bird approaching) and $\frac{dx}{dt} = 3 \, m/s$ (the speed of the fish).Substituting the values: $9 = 3 + (4/3) 	imes v_b$.$6 = (4/3) 	imes v_b$.$v_b = 6 	imes (3/4) = 4.5 \, m/s$.

$A$ fish rising vertically at the rate of $3 \, m/s$ to the surface of water observes a bird above the water,diving vertically towards it at the rate of $9 \, m/s$. The refractive index of water is $4/3$. Calculate the actual velocity (velocity $w.r.t$ ground) of the bird in $m/s$.

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