Acceleration while not in motion?

I'm really confused about another question in my assignment.

I will put the whole question as it's in two parts and I may have already made a mistake.
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Q A ball dropped from rest at 1.2m above the floor rebounds to a height of 0.65m.

Assuming the ball is not moving horizontally, calculate its velocity before it hits the floor on the way down and just after it left the floor on the way up.

(For this I used v^2 = u^2 + 2gs where, for the way down, u = 0, g = -9.8 and s = 1.2. I got -4.85 m/s.
For the way up I repeated this with u = 0, g = 9.8 and s = 0.65. I wouldn't have done this but someone in my class said to treat it like it has been dropped in reverse. So I've got 3.57 m/s).

If the ball is in contact with the floor for 0.15s determine its acceleration on the way down, while it is in contact with the floor and on the way up.
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First of all, I don't get why it has any acceleration while it is not moving for all that time. But I guess it must

I used the equation v = u + at to try and work these out, having worked out the times to be 0.25s and 0.18s respectively.

Oh, that's not what I was taught.

Have a look at this slow motion video. Notice the ball compression and the length of time it is in contact with the ground.

[video="youtube_share;zd2V4_FNMls"]http://youtu.be/zd2V4_FNMls[/video]

Both your answers are correct except that acceleration due to gravity is normally stated as +ve for the first part and deceleration due to gravity as -ve for the second part and hence the signs for the velocities are reversed.

The kinetic energy of motion as the ball hits the ground, compresses the ball and that energy is then stored in the ball in the same way as compressing a spring stores energy.

lt takes time to compress the ball and during that time, all of the molecules in the ball must reverse direction. i.e. decelerate and then accelerate upwards as the stored compression is released.

Think again:

(The equation choice is correct).

The ball is in contact with the floor for 0.15s

During which time the downwards velocity must be stopped (the ball decelerates) and then the ball accelerates upwards sufficiently to reach the rebound initial velocity.

You can assume that the time for deceleration (compression) and acceleration (rebound) are the same. i.e. 0.15/2 seconds since both are accomplished during the 0.15s the ball is in contact with the ground.

To work out the acceleration of the ball on the way down should the time be the time taken for it to travel to the floor plus half of 0.15s? i.e. 0.4s? And the same for the way up: 0.33s?

The wording of the question is the problem - its horrible and as it is rather ambiguous as to what is actually required.

"If the ball is in contact with the floor for 0.15s determine its acceleration on the way down, while it is in contact with the floor and on the way up."

Clearly, the acceleration on the way down to the point at which the ball strikes the floor is that due to gravity. i.e. 9.81m/s². At that instant the ball has a final velocity of -4.85m/s using your convention.

Again, when the ball rebounds from the instant of leaving the ground to reaching the rebound height of 0.65m, the deceleration is also that due to gravity. i.e. -9.81m/s² Again, at the instant of leaving the ground, the ball has an initial velocity of 3.57m/s

So far so good.

The problem arises with interpreting the 0.15s contact with the ground.

In that time, all parts of the ball must decelerate (against gravity) bringing it to a stop, reverse direction and then accelerate against gravity again (upwards direction).

But without dimensions of the ball, and without being told which point on the ball to take as a reference, there is no way of calculating contact acceleration and deceleration values.

So we are left with interpreting the question as a difference in values of the two known velocities.

i.e. -4.85m/s and 3.57m/s

That produces a total change in velocity magnitude of 8.42m/s.

Acceleration is rate of change of velocity which must occur in 0.15s

So the total acceleration using that assumption should be 8.42 / 0.15 = 56.13m/s²

Because of the ambiguity, I'm not sure if this is what the person who set the question intended, so perhaps we should ask for a second opinion from someone else although I'm not sure they will be able to help either.