A body of $$m$$ kg slides from rest along the curve of vertical circle from point $$A$$ to $$B$$ in friction less path. The velocity of the body at $$B$$ is: (given, $$R = 14$$ m, $$g = 10$$ m/s$$^2$$ and $$\sqrt{2} = 1.4$$)

Potential Energy at point A, $$PE_A=mgH_A\ =mgR\left(1+\sin45^{\circ\ }\right)\ =m\times\ 10\times\ \left(1+\frac{1}{\sqrt{\ 2}}\right)\ =\ 240m\ J$$

Potential Energy at point B = 0

From conservation of energy,

$$PE_A+KE_A=PE_B+KE_B$$

⇒ $$KE_B=PE_A+KE_A-PE_B\ =240m+0-0=240m$$

⇒ $$\frac{1}{2}mv_B^2=240m\ ⇒\ v_B^2\ =480$$

$$\therefore\ v_B=\sqrt{\ 480}=21.91\ $$ m/s

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