A heavy particle hanging from a string of length l is projected horizontally with speed $\sqrt{\mathrm{gl}}$ . The speed of the particle at the point where the tension in the string equals weight of the particle is

$\mathrm{T}-\mathrm{mg} \cos \mathrm{\theta } = \frac{{\mathrm{mv}}^2}{\mathrm{R}}$

Given T = mg

$\mathrm{mg}-\mathrm{mg} \cos \mathrm{\theta } = \frac{{\mathrm{mv}}^2}{\mathrm{R}}$

$\mathrm{g}(1-\cos \mathrm{\theta }) = \frac{{\mathrm{v}}^2}{\mathrm{R}}$--------------(i)

C.O.M.E. at A and B ; $∆\mathrm{K} + ∆\mathrm{U} = 0$

$(\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}{\mathrm{mu}}^2) + \mathrm{mg}(\mathrm{R}-\mathrm{R} \cos \mathrm{\theta }) = 0$

$\Rightarrow {\mathrm{v}}^2-{\mathrm{u}}^2 = -2\mathrm{gR}(1-\cos \mathrm{\theta })$

$\Rightarrow {\mathrm{v}}^2-{\left(\sqrt{\mathrm{gl}}\right)}^2 = -2{\mathrm{v}}^2$

$3{\mathrm{v}}^2 = \mathrm{gl} \Rightarrow \mathrm{v} = \sqrt{\frac{\mathrm{gl}}{3}}$