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(B) At the lowest position,the velocity is $v$. The acceleration is purely centripetal,given by $a_{low} = \frac{v^2}{\ell}$.At the extreme position,t

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$2 	an ^{-1}\left(\frac{1}{2}ight)$

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(B) At the lowest position,the velocity is $v$. The acceleration is purely centripetal,given by $a_{low} = \frac{v^2}{\ell}$.At the extreme position,the velocity is zero. The acceleration is purely tangential,given by $a_{ext} = g \sin 	heta$.By conservation of energy between the lowest point and the extreme point:$\frac{1}{2} mv^2 = mg \ell(1 - \cos 	heta) \Rightarrow \frac{v^2}{\ell} = 2g(1 - \cos 	heta)$.Given that the magnitudes of acceleration are equal:$a_{low} = a_{ext} \Rightarrow \frac{v^2}{\ell} = g \sin 	heta$.Substituting the expression for $\frac{v^2}{\ell}$:$2g(1 - \cos 	heta) = g \sin 	heta \Rightarrow 2(1 - \cos 	heta) = \sin 	heta$.Using half-angle identities $1 - \cos 	heta = 2 \sin^2(	heta/2)$ and $\sin 	heta = 2 \sin(	heta/2) \cos(	heta/2)$:$2(2 \sin^2(	heta/2)) = 2 \sin(	heta/2) \cos(	heta/2)$.Dividing by $2 \sin(	heta/2)$ (assuming $	heta 
eq 0$):$2 \sin(	heta/2) = \cos(	heta/2) \Rightarrow 	an(	heta/2) = \frac{1}{2}$.Therefore,$	heta = 2 	an^{-1}\left(\frac{1}{2}ight)$.

$A$ ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle $(\theta)$ of thread deflection in the extreme position will be :

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