This is a conservation of energy and momentum problem. The bob's potential energy converts to kinetic energy at the bottom: $\frac{1}{2}m{v}^2=mg\ell$, so $v=\sqrt{2g\ell }$. At the bottom, an inelastic collision occurs. Momentum is conserved: $mv=2m{v}_{\text{f}}$, giving final velocity $v_{\text{f}}=\frac{v}{2}$. The new system (mass 2m) rises, converting kinetic energy to potential energy: $\frac{1}{2}\left(2m\right){v_{\text{f}}}^2=2mgh$. Solving for height $h=\frac{v^2}{8g}=\frac{\ell }{4}$. The maximum angle θ from vertical is found from geometry: $h=\ell (1-\cos \theta )$. Substituting gives $\cos \theta =\frac{3}{4}$, so $\theta ={\cos }^{-1}\left(\frac{3}{4}\right)$.

Final Answer: ${\cos }^{-1}\left(\frac{3}{4}\right)$