Apply the kinematic equations to free-fall motion using a constant acceleration of 9.8 m/s^2 due to gravity.
Near Earth's surface, any object moving through the air under gravity alone (ignoring air resistance) accelerates downward at a constant 9.8 m/s^2 -- a value physicists call "g." This is just a special case of last week's constant-acceleration kinematics, with a = -9.8 m/s^2 if we choose "up" as the positive direction, whether the object is rising, falling, or momentarily paused at the very top of its path. Famously, this acceleration does not depend on mass: a bowling ball and a tennis ball, dropped together with no air resistance, hit the ground at the exact same instant.
Free fall has a useful symmetry when an object returns to its starting height: the time spent rising equals the time spent falling, and the speed at any given height is the same on the way up as on the way down (same magnitude, opposite direction). At the very peak of a toss straight up, velocity is momentarily zero -- but acceleration is NOT zero; gravity never stops pulling, which is exactly why the object does not stay at the peak.
Mastery looks like: They apply the kinematic equations to a free-fall scenario with a = -9.8 m/s^2, correctly finding time, height, or speed as needed.
Common stumbles: Believing heavier objects fall faster; assuming that because velocity is zero at the peak, acceleration must be zero there too.