\( \Rightarrow \frac{1}{2} \mathrm{mv}^2=\mathrm{mg} \ell(1-\cos \theta)\)

\( \Rightarrow \frac{\mathrm{v}^2}{\ell}=2 \mathrm{~g}(1-\cos \theta)\)

Acceleration at lowest point \(=\frac{\mathrm{v}^2}{\ell}\)

Acceleration at extreme point \(=g \sin \theta\)

Hence, \(\frac{\mathrm{v}^2}{\ell}=\mathrm{g} \sin \theta\)

\( \therefore \sin \theta=2(1-\cos \theta) \)

\( \Rightarrow \tan \frac{\theta}{2}=\frac{1}{2} \Rightarrow \theta=2 \tan ^{-1}\left(\frac{1}{2}\right)\)