Physics - W.E.P. Question with Solution | TestHub

See the diagramLet the velocity be $v$ when the body leaves the surface. From free body diagram, $\frac{m{v}^2}{R}=mg\cos \theta$. $\Rightarrow v^2=Rg\cos \theta$ ...(i) Again from work energy principle, change in $\mathrm{KE}=$ work done $\Rightarrow \frac{1}{2}m{v}^2-0=mg[R-R\cos \theta ]$ $v^2=2gR(1-\cos \theta )$ ...(ii) From Eqs. (i) and (ii) $\begin{array}{rl}Rg\cos \theta & =2gR(1-\cos \theta )\\ 3gR\cos \theta & =2gR\\ \Rightarrow \cos \theta & =\frac{2}{3}\Rightarrow \theta ={\cos }^{-1}\left(\frac{2}{3}\right)\end{array}$