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?
$$\left(g = 10\ \frac{\text{m}}{\text{s}^2},\ \cos 53^\circ = 0.6\right)$$
[Figure: Pendulum of length $1\ \text{m}$ released from point $A$ at $53^\circ$ from vertical](Air resistance is neglected.)
- [(1)] $60$ (2) $45$
- [(3)] $37$ (4) $30$