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A particle of mass m is projected with speed (Rg/4)^{1/2} from top of a smooth hemisphere. If the particle starts slipping from the highest point, then the horizontal distance between the point, then horizontal distance between point where it leaves contact with sphere and the point at which the body was placed is

Answer is R(7)^{1/2 }/4 can you explain how ?

when the particle is about to leave the surface normal force on it is zero.so centripetal force is provided by component of weight towards centre.

${f}_{c}=\frac{m{v}^{2}}{r}\phantom{\rule{0ex}{0ex}}mg\mathrm{cos}\theta =\frac{m{v}^{2}}{r}\phantom{\rule{0ex}{0ex}}g\mathrm{cos}\theta =\frac{{v}^{2}}{R}.....1\phantom{\rule{0ex}{0ex}}fromconservationofmechanicalenergy\phantom{\rule{0ex}{0ex}}mgR+\frac{1}{2}m{{v}_{0}}^{2}=mgR\mathrm{cos}\theta +\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}gR+\frac{1}{2}\frac{gR}{4}=gR\mathrm{cos}\theta +\frac{1}{2}gR\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}\frac{9}{8}=\frac{3}{2}\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{3}{4}\phantom{\rule{0ex}{0ex}}horizontaldis\mathrm{tan}cefrompointofprojection=R\mathrm{sin}\theta =R\times \frac{\sqrt{7}}{4}$

Regards

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