What happens to a tennis ball in free-fall when it is dropped?
Let us assume you are dropping the ball from a height H. At the instant the ball leaves your hand it has no velocity. But a force of gravitational attraction is acting on the ball that makes it accelerate downwards at 9.8 m/s^2. This number may vary slightly depending of the altitude and the resistance due to air, but for general purposes we can take the figure to be 9.8 m/s^2.
The velocity of ball t seconds after it has left your hand is equal to 9.8*t m/s.
The distance the ball travels in t seconds is equal to (1/2)*9.8*t^2 m
At first the ball will accelerate towards the ground at about 9.8 metres per second squared. This number gets less as the ball gets faster due to air drag but not by much if you drop it from head height.