Answer
The vertical component $a_y$ of the particle’s acceleration is greatest in magnitude at $\theta = 90^{\circ}$ and $\theta = 270^{\circ}$
Work Step by Step
The magnitude of acceleration is constant and $~~a = \frac{v^2}{r}$
The acceleration vector is always directed toward the center of the circle.
At $\theta = 90^{\circ}$, the acceleration vector is directed straight down so the magnitude of $a_y$ is a maximum.
At $\theta = 270^{\circ}$, the acceleration vector is directed straight up so the magnitude of $a_y$ is a maximum.
The vertical component $a_y$ of the particle’s acceleration is greatest in magnitude at $\theta = 90^{\circ}$ and $\theta = 270^{\circ}$
