A ball is dropped from a height of 20 m . A fish sees the ball 12.8 m above the surface of water. What is its apparent speed with which its see the ball.?

Question

Answer

Fish will see the image of ball after reraction through water surface...

using

n2/v - n1/u = n2-n1/R

R for plane surface is infinity & n2(refrative index of water) , n1(refractive index of air)

4/3v - 1/u = 0

v = 4u/3

now differentiating eq 1 wrt t

(velocity of image)dv/dt = 4/3.(du/dt) (velocity of object)

now when ball has fallen 12.8m then its velocity is U..

so velocity of ball as seen by fish is (4/3) .U

V = 4/3. (U) =4/3 (sqrt2g(H-h)) (change in height (H-h) = 7.2m)

=4*12/3=16m/s

Answer

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