Sign of acceleration due to gravity problem

1. The problem statement, all variables and given/known data

A ball is dropped and rebounds vertically to less than the original height. For this first bounce only, sketch the graph of
(a) velocity against time (2) (b) acceleration against time (2)



2. Relevant equations

The solution I have for the acceleration time graph has g as -9.81 as the ball descends and -9.81 as the ball is ascending.


3. The attempt at a solution




The graph also had velocity increasing with a uniform acceleration from zero down to its maximum value just before it hits the surface then a change of sign and it is now positive with the same velocity as just before impact. This then decreases with a uniform deceleration until the velocity reaches zero at the maximum bounce hight.

Is g always minus 9.81 because during descent it is in the negative direction in my coordinate system but accelerating, then negative for the ascent because it is in the positive direction in my coordinate system but decelerating?

In most of my SUVAT problems g was simply positive 9.81 when descending and -9.81 when ascending

any help appreciated

 

Hi thanks for the help

should I be thinking if the velocity is ever zero during its contact with the floor?
I assume in the moment of changing direction it is stationary and must have a velocity of zero and hence cross the x axis at some point between the start of contact with the floor and then leaving the floor as it returns upwards?

my other confusion comes from the fact that the minus sign seems to convey information about whether the change in velocity (acceleration) is increasing or decreasing in magnitude and also information about its direction in my defined coordinate system. This then gives us a situation that g is always minus in this problem, when the magnitude of the velocity is increasing as it decends and when the magnitude of the velocity is decreasing as it ascends.

Thanks