A circle and a point $P$ higher than the circle lie in the same vertical plane. A particle moves along a straight line under gravity from $P$ to a point $Q$ on the circle. Given that the distance travelled from $P$ in time $t$ is equal to $\dfrac{1}{2}gt^2 \sin{\alpha}$, where $\alpha$ is the angle of inclination of the line $PQ$ to the horizontal, give a geometrical characterization of the point $Q$ for which the time taken from $P$ to $Q$ is a minimum.
1. **Define the problem geometrically:**
   - Let \( O \) be the center and \( r \) the radius of the given circle.
   - Let \( P \) be a point higher than the circle in the same vertical plane.
   - A particle moves along a straight line under gravity from \( P \) to a point \( Q \) on the circle.
   - The distance traveled from \( P \) in time \( t \) is given by \( \frac{1}{2}gt^2 \sin{\alpha} \), where \( \alpha \) is the angle of inclination of the line \( PQ \) to the horizontal.

2. **Express the time \( t \) in terms of \( d_1 \) and \( \alpha \):**
   - The distance \( d_1 = PQ \).
   - The time \( t \) is given by:
     \[
     t = \sqrt{\frac{2d_1}{g \sin \alpha}}
     \]
   - To minimize \( t \), we need to minimize \( \sqrt{\frac{d_1}{\sin \alpha}} \), which is equivalent to minimizing \( \frac{d_1}{\sin \alpha} \) since the square root function is increasing.

3. **Use the power of a point theorem:**
   - The power \( p \) of the point \( P \) to the circle \( O \) is constant for any angle \( \alpha \) and is given by:
     \[
     p = d_1 d_2 = (OP - r)(OP + r) = OP^2 - r^2
     \]
   - Here, \( d_2 = PR \) where \( R \) is another point of intersection of the line \( PQ \) with the circle.

4. **Minimize the expression \( \frac{d_1}{\sin \alpha} \):**
   - We have:
     \[
     \frac{d_1}{\sin \alpha} = \frac{p}{d_2 \sin \alpha}
     \]
   - This expression is minimized when \( \frac{1}{d_2 \sin \alpha} \) is minimized, or equivalently when \( d_2 \sin \alpha \) is maximized.

5. **Maximize \( d_2 \sin \alpha \):**
   - Let the vertical line through \( P \) intersect the horizontal line through \( R \) at a point \( S \).
   - The distance \( PS = d_2 \sin \alpha \).
   - This distance is maximized when \( R \) is at the lowest point \( B \) of the circle, i.e., the bottom intersection of the vertical line through the center \( O \) with the circle.

6. **Determine the desired point \( Q \):**
   - The desired point \( Q \) is the other intersection of the circle \( O \) with the line \( PB \).

The final answer is the point \( \boxed{ Q } \) is the other intersection of the circle \( O \) with the line \( PB \).