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A small ball is thrown from the edge of one bank of a river of width 100m to just reach the bank. The ball was thrown in the vertical plane (which is also perpendicular to the banks) at an angle 37^(@) to the horizonatal. Taking the starting the point as the origin O, verection as positeve y-axis and horizonrtal line passing through the point O and perpenducular to the bank as x-axis find: (a) Equation of trahectory of the image formed by the water surface (water surface is at the level y=0) (b) instantaneous velocity of the image formed due to refraction. [Use g=10 m//s^(2),R.I. of water =4//3] |
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Answer» `therefore` apparent trajectory or image of the trahectory will have `'x'` unchanged but `'y'` multipled by a factor of `R.I.=4//3` `therefore` Equation is `y'=(4)/(3)y=(4)/(3)xtan 37^(@)(a-(x)/(100))` or `y=(x(x^(2))/(100))` `(u^(2)sin(2xx37^(@)))/(G)=100 rArru=25sqrt(5/3)` (b) Similarly `'x'` COMPONENT of velocity will remain unchanged, but the `'y'` component of velocity will be multiplied by a factor of `R.I.=4//3` or at any time `'t'`. `v(t)_(image)=u cos 37^(@)i+(4)/(3)(u sin 37^(@)-gt)j=25sqrt(5/3)xx(4)/(5)i+(4)/(3)[25sqrt(5/3)xx(3)/(5)-10t]j` `v(t)_("image")=20sqrt(5/3)i+20[sqrt(5/3)-(2)/(3)t]j` |
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