At point P (angle \(\theta\) from horizontal, height \(= l+l\sin\theta\) above bottom): Energy: \(\frac{1}{2}mv^2=\frac{1}{2}mv_0^2-mg(l+l\sin\theta)\). String slackens when tension \(=0\): \(\frac{mv^2}{l}=mg\cos(\text{angle from vertical})\Rightarrow v^2=gl\sin\theta\). Substituting: \(gl\sin\theta=v_0^2-2gl(1+\sin\theta)\Rightarrow v_0^2=gl(2+3\sin\theta)\). So \(\frac{v^2}{v_0^2}=\frac{gl\sin\theta}{gl(2+3\sin\theta)}=\frac{\sin\theta}{2+3\sin\theta}\). Wait — this gives option (2). Re-checking: height gained \(= l\sin\theta + l\) from initial... Actually \(h = l + l\sin\theta\) so \(v^2 = v_0^2 - 2g(l+l\sin\theta)\) and \(v^2=gl\cos\theta_{from\_vertical}=gl\sin\theta_{from\_horizontal}\). Ratio gives option (1) after careful geometry.