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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» Velocity of the body at height h, `v_h=sqrt(2g(H-h))`, horizontally (from the figure given in the problem). Time taken in falling through the distance h. `t=sqrt((2h)/(g))` (as initial vertical component of the velocity is zero.) Now `s=v_ht=sqrt(2g(H+h))xxsqrt((2h)/(g))=sqrt(4(Hh-h^2))` For `s_(max)`, `(d)/(ds)(Hh-h^2)=0`, which yields `h=H/2` Putting this value of h in the expression obtained for s, we get, `s_(max)=H` |
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