# Sign of acceleration due to gravity problem

## Homework Statement

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)

## Homework 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.

## 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

The graph of velocity (with const acceleration ) is given by v = u+ft so is a staight line in each segment.

Yes the acceleration f = acceleration due to gravity =g and g is always negative if you reckon up is positve.

When you first drop the ball u = 0 and it falls, accelerating downwars along the line v = -gt (ie through the origin) until it reaches vf at the floor. (vf here is negative).
For an elastic bounce it rebounds with initial velocity -vf ie upwards along the line

v = (+vf) -gt

Do you think it crosses the axis as you have shown?

Hi thanks for the help

"Do you think it crosses the axis as you have shown?"
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

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?
Difficult to tell on your graph but you need to consider how long the ball spends in contact with the floor when it bounces.

If we consider it instantaneous this implies a change of velocity from -vf to +v(f - δvf) in zero time ie as a vertical line connecting these two points. If it is not instantaneous but takes place over a short period δt then the connecting line will be slightly slanted forward as you have shown.

The return velocity after bouncing is (vf-δvf) to allow for the condition that it does not rebound to the initial height. This then generates a decaying sawtooth, rahter than one of constant amplitude.

Gravity is always acting, so the acceleration due to gravity is always acting and negative, as you have drawn. Superimposed on this is a positive pulse, of duration δt, again as you have drawn.

You can again use v=u+ft, this time to evaluate the positive acceleration pulse for time δt, if we assume a constant acceleration necessary for u = -vf; v = +(vv-δvf) can you rearrange this formula to see how big a positive pulse is rquired?

I suggest the acceleration graph is drawn underneath or on top of the velocity graph to show the temporal coincidence of the pulse and the upstroke of the sawtooth.

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