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(C) At point $A$ (extreme position),the velocity is zero. The acceleration is purely tangential,given by $a_A = g \sin 	heta$.At point $B$ (mean posi

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$cos^{-1}(3/5)$

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(C) At point $A$ (extreme position),the velocity is zero. The acceleration is purely tangential,given by $a_A = g \sin 	heta$.At point $B$ (mean position),the acceleration is purely radial (centripetal),given by $a_B = \frac{v^2}{l}$.Using the conservation of energy between $A$ and $B$,the change in potential energy equals the change in kinetic energy: $mgl(1 - \cos 	heta) = \frac{1}{2}mv^2$,which implies $v^2 = 2gl(1 - \cos 	heta)$.Substituting this into the expression for $a_B$: $a_B = \frac{2gl(1 - \cos 	heta)}{l} = 2g(1 - \cos 	heta)$.Given that $a_A = a_B$,we have $g \sin 	heta = 2g(1 - \cos 	heta)$.Dividing by $g$: $\sin 	heta = 2 - 2 \cos 	heta$.Squaring both sides: $\sin^2 	heta = (2 - 2 \cos 	heta)^2 \Rightarrow 1 - \cos^2 	heta = 4(1 - \cos 	heta)^2$.$1 - \cos^2 	heta = 4(1 - 2 \cos 	heta + \cos^2 	heta) \Rightarrow 1 - \cos^2 	heta = 4 - 8 \cos 	heta + 4 \cos^2 	heta$.$5 \cos^2 	heta - 8 \cos 	heta + 3 = 0$.Solving the quadratic equation for $\cos 	heta$: $\cos 	heta = \frac{8 \pm \sqrt{64 - 4(5)(3)}}{2(5)} = \frac{8 \pm \sqrt{64 - 60}}{10} = \frac{8 \pm 2}{10}$.This gives $\cos 	heta = 1$ (which corresponds to $	heta = 0$,the mean position) or $\cos 	heta = 6/10 = 3/5$.Thus,$	heta = \cos^{-1}(3/5)$.

$A$ ball attached to one end of a string swings in a vertical plane such that its acceleration at point $A$ (extreme position) is equal to its acceleration at point $B$ (mean position). The angle $\theta$ is :-

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