A pendulum bob has a speed

Applying work energy theorem :

$\mathrm{w}_{\mathrm{g}}+\mathrm{w}_{\mathrm{T}}=\Delta \mathrm{K}$

$-\mathrm{mgl}\left(1-\cos 60^{\circ}\right)=\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{mu}^{2}$

$\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{gl}\left(1-\cos 60^{\circ}\right)$

$\mathrm{v}^{2}=9-2 \times 10 \times 0.5\left(\frac{1}{2}\right)$

$\mathrm{v}^{2}=4$

$\mathrm{v}=2 \mathrm{~m} / \mathrm{s}$