$A$ ball is thrown from a rooftop at an angle of $45^\circ$ above the horizontal. It hits the ground a few seconds later. At what point during its motion does the ball have $(a)$ greatest speed, $(b)$ smallest speed, and $(c)$ greatest acceleration? Explain.

(N/A) Consider the given figure in which a ball is projected from point $O$ and follows the path $O-A-B-C$ to reach the ground at $C$.
$(a)$ The speed of the projectile is given by $v = \sqrt{v_x^2 + v_y^2}$. Since the horizontal component $v_x$ remains constant and the vertical component $v_y$ increases in magnitude as the ball falls below the level of projection, the speed is greatest just before the ball hits the ground at point $C$.
$(b)$ The speed is smallest at the highest point of the trajectory (point $A$), where the vertical component of velocity $v_y$ is zero, and the speed is equal to the horizontal component $v_x = v_0 \cos 45^\circ$.
$(c)$ The acceleration of the ball is due to gravity, which acts vertically downwards throughout the motion. Therefore, the acceleration is constant and equal to $g$ at all points during the motion.