\textbf{164-} In the figure below, a pendulum ball is released from point $A$ and passes through the lowest point of the path with speed $V$. When the ball's speed reaches $\dfrac{\sqrt{2}}{2}\ V$, what angle does the string make with the vertical?
$$\left(g = 10\ \frac{\text{m}}{\text{s}^2},\ \cos 53^\circ = 0.6\right)$$
\textit{[Figure: Pendulum of length $1\ \text{m}$ released from point $A$ at $53^\circ$ from vertical]}
(Air resistance is neglected.)
\begin{itemize}
\item[(1)] $60$ \hfill (2) $45$
\item[(3)] $37$ \hfill (4) $30$
\end{itemize}