Problem source

A particle $P$ is projected, from the lowest point, along the smooth inside surface of a fixed sphere with centre $O$. It leaves the surface when $OP$ makes an angle $\theta$ with the upward vertical. Find the smallest angle that must be exceeded by $\theta$ to ensure that $P$ will strike the surface below the level of $O$. 
\textit{You may find it helpful to find the time at which the particle strikes the sphere.}

Solution source

\begin{center}
        \begin{tikzpicture}[scale = 2]

        \coordinate (O) at (0,0);
        \coordinate (C) at ({4/5}, {3/5});
        \coordinate (X) at (0, 1);

        \draw[dotted] (O) -- (C);
        \draw[dotted] (O) -- (X);

        \filldraw (O) circle (1pt);
        \filldraw (C) circle (1pt);

        \draw (O) circle ({1});

        \draw[-latex, blue, ultra thick] (C) -- ($(O)!0.5!(C)$) node[left] {$R$};
        \draw[-latex, blue, ultra thick] (C) -- ++(0, -0.5) node[below] {$mg$};
        
        \pic [draw, angle radius=1cm, angle eccentricity=1.25, "$\theta$"] {angle = C--O--X};
        
        \draw[-latex, red, ultra thick] (1.5,0) -- (1.2,0);
        \node[right, red] at (1.5,0) {$0$ G.P.E};
        
    \end{tikzpicture}
\end{center}

\begin{align*}
%\text{COE}: && \frac12 m u^2 - mga &= \frac12mv^2 + mga\cos \theta \\
\text{N2}(\swarrow): && R+mg\cos\theta &= \frac{m v^2}{a} \\
R = 0: && v^2 &= ag\cos \theta \\
\end{align*}

So the particle will become a projectile moving tangent to the circle with $v^2 = ag \cos \theta$.

Therefore the velocity will be $\displaystyle \sqrt{ag \cos \theta}\binom{-\cos \theta}{\sin \theta}$.

We have:

\begin{align*}
&& \mathbf{s} &= a\binom{\sin \theta}{\cos \theta}+\sqrt{ag \cos \theta}\binom{-\cos \theta}{\sin \theta} t + \frac12 \binom{0}{-g} t^2 \\
\Rightarrow && a^2 &= \mathbf{s} \cdot \mathbf{s} \\
&&&= a^2 + ag\cos \theta t^2 + \frac1{4} g^2t^4 -ag \cos \theta t^2 - \sqrt{ag \cos \theta} \sin \theta g t^3 \\
\Rightarrow && 0 &= \frac14 g t - \sqrt{ag \cos \theta} \sin \theta  \\
\Rightarrow && t &= \frac{4\sqrt{a g \cos \theta} \sin \theta}{g}  
\end{align*}

At this time, the vertical position will be:

\begin{align*}
&& s_y &= a \cos \theta + \sqrt{ag \cos \theta} \sin \theta \frac{4\sqrt{a g \cos \theta} \sin \theta}{g}   - \frac12 g \left ( \frac{4\sqrt{a g \cos \theta} \sin \theta}{g}  \right)^2 \\
&&&= a \cos \theta + 4a\cos \theta \sin^2 \theta - 8a\cos \theta \sin^2 \theta \\
&&&= a \cos \theta - 4 a \cos \theta \sin^2 \theta \\
&&&= a \cos \theta (1-4 \sin^2 \theta) \\
\underbrace{\Rightarrow}_{s_y < 0} && 0 &> 1 - 4 \sin^2 \theta \\
\Rightarrow && \sin\theta &> \frac12 \\
\Rightarrow && \theta & > \frac{\pi}{6}
\end{align*}