A ball is dropped from rest near the surface of Earth as shown in Animation 1. Its motion is described as free fall (position is given in meters and time is given in seconds). If the +y direction is defined to be upward, then the ball's acceleration is constant and has a value of -9.8 m/s². If the +y direction is defined to be downward, then the acceleration is +9.8 m/s².

Animation 2 shows the ball thrown upward such that it travels upward, reaches some maximum height, and falls back to your hand at the same height at which is was first thrown. Consider the motion of the ball only while it is in the air and not in your hand (if it's in your hand, it's not considered free fall). The green vector shown is the velocity vector of the ball, v_y. When the ball is at its peak? What is v_y?

Look at the graphs. At the peak of its motion, the ball goes from upward (a positive velocity) to downward (a negative velocity). It changes velocity smoothly, and the velocity must go through zero at this turnaround point. The acceleration is the change in velocity, which is constant throughout the motion at -9.8 m/s². You can measure the velocity at two different times by clicking in the v vs. t graph to see that Δv/Δt is constant.

Animation 3 shows the ball thrown downward. Notice that while the ball travels farther (note that it is off screen for most of its motion) and moves much faster than when it is dropped from rest, the slope of the velocity vs. time graph is still -9.8 m/s² as shown by the acceleration vs. time graph.