Have I sufficiently demonstrated the answer? A point is reached on an inclined plane that lies on a circle in the time take to fall directly.

Here is my calculation.

When $frac{1}{2}alphasin A t^2=OQ$, $frac{1}{2}alpha t^2=OP$

Therefore, $frac{OQ}{sin A}=OP$ and $frac{OQ}{OP}=sin A$

If we draw a line and name it $QP$ we draw a triangle inside the circle that satisfies the calculation $frac{OQ}{OP}=sin A$.

I don’t have much experience with mathematics so is this answer sufficient to show that the point $Q$ lies on the circle?