Question

Can anyone explain this velocity-time graph of a bouncing ball at Fig.2.2?

Answer 1

Well firstly we must understand that in this graph, the positive velocity is actually downward, which you will see in a second.
So look at the start as the ball being dropped from some height, as it falls it gains velocity until it strikes the ground and suddenly the downward kinetic energy moves to elastic potential energy in the ball which is then pushed out towards the ground into kinetic energy. this action is the ball bouncing! so at that sharp slant, it is actually bouncing off of he ground, 1.15 is when all the energy is stored in the ball, and then it is pushed out and the ball rises at a similar speed to what it was incoming at, and it is now traveling upward (some velocity is lost to thermal and sound ect.) so now it is traveling upwards i think you can figure out the rest, just remember in this case negative velocity is actually upward movement, and we have a certain constant g that is effecting our ball through out this process so the part after the bounce should be easy for you to explain either way would like to see some work from you too!

Answer 2

So the graph below the x-axis is the ball travelling upwards after its bounce, then its v=0ms-1 in which its height is maximum then the ball is falling downwards which is the graph above the x-axis.
@memman