InterviewSolution
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InterviewSolution
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A small disc A slides down with initial velocity equal to zero from the top of a smooth hill of height H having a horizontal portion. What must be the height of the horizontal portion h to ensure the maximum distance s covered by the disc? What is it equal to? |
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Answer» Correct Answer - B In order to obtain the velocity point (B), we apply the law of conservation of energy. So, Loss in `PE`=Gain in `KE` `m g (H-h) =1/2mv^(2)` `:. v=(sqrt[2g(H-h)])` Further `h=1/2gt^(2)` `:. t=sqrt((2h)/g)` Now, `s=v xx t =sqrt[[2g(H-h)]] xx sqrt((2 h//g))` or `s=sqrt[[4h(H-h)]]` .....(i) For maximum value of s, `(ds)/(dh)=0` `:. 1/(2sqrt[[4h(H-h)]]xx4(H))=0` or `h=H/2` Substituting `h=H//2` in E.q. `(1)`, we get `s=sqrt[[4(h//2)(H-//2)]]=sqrt(H^(2))=H` |
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