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A particle slides down the surface of a smooth fixed sphere of radius R starting from rest at the highest point B. Particle leaves the sphere at some point and then strikes the horizontal plane passing through the lowest point A of the sphere at point P. The distance AP is given by AP=n/27[sqrt(n)+4sqrt(2)]R where 'n' is an integer, find n (Take g=10m//s^(2)) |
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Answer» `y=(v SIN theta)t+1//2 "gt"^(2)=R(1+costheta)` ……..(1) and `x=(v cos theta)t` ………(2) and `v^(2)=2/3 gR`…..(3) On soving `AP= Rsin theta +x=5/27 (sqrt(5)+4sqrt(2))R` 
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