A pendulum bob has a speed of 3m s ^ - 1 at its lowest position....

Applying work energy theorem:

${w}_{g} + {w}_{T} = \DeltaK$

$- mgl\left(1 - \cos 60 °\right) = \dfrac{1}{2}m{v}^{2} - \dfrac{1}{2}m{u}^{2}$

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

${v}^{2} = 9 - 2 \times 10 \times 0 . 5\left(\dfrac{1}{2}\right)$

${v}^{2} = 4$

$v = 2m{s}^{- 1}$