An "out of the box" lesson:
Consider the following vertical, one dimensional motion:
y(t) = 200t/3 - 5t^2 + (t^3)/6 -(t^4)/480
Take two derivatives and plot y(t) , v(t) , and a(t) from t=0 to 40.
Observe that the thing seems to "hang" momentarily in the air:
At t=20 the object has reached the top of its path.
It then turns around and falls.
But at t=20, both the velocity and the acceleration are zero!
This should disabuse one of (sometimes spoken, sometimes implied) arguments
that the acceleration cannot be zero at a turnaround point.
PS. Follow this presentation with the observation that there is nothing
inherently special about the turnaround point in a free fall vertical toss
as observed from the ground. Go to the inertial frame of a vertically
moving helicopter and you can make any event in the ball's travel the
turnaround point (or eliminate the turning point altogether).