27.5 As shown in the figure, in the vertical coordinate plane $x O y$, the line $l$ passes through the origin $O$, and $l$ is in the first and third quadrants, with the angle between $l$ and the $x$-axis being $\alpha\left(0^{\circ}<\alpha<90^{\circ}\right)$. There is a point mass (neglecting the size of the point mass) at position $A$ directly above the $O$ point on the $y$-axis. The point mass starts from rest and slides along a straight path to point $B$ on $l$ and stops. Let $\angle B A O=\theta$. When the time taken for the point mass to slide to $l$ is the shortest, $\theta$ equals
(A) $0^{\circ}$.
(B) $\frac{\alpha}{4}$.
(C) $\frac{\alpha}{2}$.
(D) $\alpha$.
[Solution] As shown in the figure, let $O A=a$, and the gravitational acceleration be $g$. Clearly, $\angle A O B=90^{\circ}-\alpha$.
By the sine rule, we have
$$
\frac{A B}{\sin \left(90^{\circ}-\alpha\right)}=\frac{a}{\sin \left(180^{\circ}-90^{\circ}+\alpha-\theta\right)},
$$

which simplifies to
$$
A B=\frac{a \cdot \cos \alpha}{\cos (\theta-\alpha)} .
$$

Since the component of gravitational acceleration $g$ in the direction of $A B$ is $g \cdot \cos \theta$, according to the physics formula, the time taken for the particle to slide from $A$ along the line segment $A B$ to $l$ is
$$
t=\sqrt{\frac{2 \cdot A B}{g \cdot \cos \theta}}=\sqrt{\frac{2 a \cos \alpha}{g \cdot \cos \theta \cdot \cos (\theta-\alpha)}},
$$

where $\alpha, a, g$ are constants.
To minimize $t$, we only need to maximize $\cos \theta \cdot \cos (\theta-\alpha)$.
By $\cos \theta \cdot \cos (\theta-\alpha)=\frac{1}{2}[\cos (2 \theta-\alpha)+\cos \alpha]$,
we know that $\cos \theta \cdot \cos (\theta-\alpha)$ is maximized when $2 \theta-\alpha=0$.
Therefore, when $\theta=\frac{\alpha}{2}$, the time taken for the particle to slide is minimized.
Hence, the answer is $(C)$.