This is a graph of velocity versus time, so the slope of the graph at any point gives the change in velocity over time, or dv/dt which is acceleration. Acceleration occurs due to the application of force to a mass (f=ma, or a=f/m).

Any change in the slope of the graph must occur due to a change in (dv/dt), which is a change in acceleration, and hence a change in the applied force. The only forces at work here are gravity and the "spring" like force caused by deformation of the ball. The changes at 1s and 1.25s are due to the addition and removal of the "spring" force from the deforming ball as it strikes and then leaves the ground. Gravity applies throughout the graph. Both these forces are "constant" where applied, giving uniform acceleration and hence a straight line dv/dt graph with some slope.

Velocity-time graphs can be slightly confusing in that a change in direction of motion, something we would see as a major event, is indicated only where the line crosses zero. If this change of direction is due to constant acceleration, like a ball at the top of its arc or at the transition of a spring-like deformation from compression to expansion, this occurs in the middle of a line of constant slope (2.25s and 1.125s above), and seems unremarkable. Crossing zero is always a point of potential interest in a graph.