The graph of velocity against time which which might have been expected is shown on the left.

Whilst in the air the acceleration of the ball is $g=+9.8\,\rm m\,s^{-2}$ and this is the gradient of the velocity time graph.
There is a problem with this graph in that at time $2\,\rm s$ the gradient of the graph is infinite which implies that the acceleration is infinite.
There cannot be an infinite acceleration as this would imply that an infinite force is acting on the ball when it is in contact with the ground.
Expanding the time axis in the region of the red rectangle the graph might look like that on the right.
Here it was assumed that the acceleration of the ball when in contact with the ground for $\frac{2}{1000}^{\rm th}$ of a second was constant and its value can be worked out as being $-19,600\,\rm m\,s^{-2}$.
This means that on the time scale of the left graph the line at time $2$ seconds is effectively vertical.
Note that the graph of the right still cannot be correct as, as drawn, the gradient is not defined at times $1.999\,\rm s$ and $2.001\,\rm s$.
So what might the graph actually look like around those times?