When a superball rolls out the window...

The question said the building is very tall, and that means the ball will reach it's terminal velocity: The air drag (Fa) is proportional to the ball's velocity and directed in the opposite direction. So, when it reach the same magnitude of the gravitational force (Fb), the resultant force on the ball will be Fa-Fb=0. So, the ball will be in a constant velocity (V). When it hits the ground, the velocity will keep it's magnitude (thanks to the perfectly elastic collision), and the ball will go in the opposite direction. The same reversal of direction will happen to the air drag, but obviously the gravitational force won't change. So, the magnitude of the resultant force in the ball will be: F=fa+Fb=2Fb=2 m 9.8=19.6 m, and the acceleration is therefore $19.6~m/s^2$ .

Since the ball is dropped from a very high place, the time it takes to hits the ground is long enough, so it reaches its terminal velocity, as the air resistance is equal to the weight of the ball itself. After the collision with the ground, the velocity of the ball changes its direction, and the total force acts on the ball is the sum of the ball's weight and the air resistance - which is also equal to the ball's weight. So the acceleration of the ball right after the collision is $a=2g=19.6~m/s^2$ .