The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle $\theta$ and then released. The bob will then pass through its equilibrium position with a speed v, where v equals
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If suppose bob rises up to a height h as shown then after releasing potential energy at extreme position becomes kinetic energy of mean position
$\Rightarrow mgh = \frac{1}{2}mv_{\max}^2$ $\Rightarrow V_{\max} = \sqrt{2gh}$
Also, from figure $\cos \theta = \frac{l - h}{l}$
$\Rightarrow h = l(1 - \cos \theta)$
So, $v_{\max} = \sqrt{2gl(1 - \cos \theta)}$
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