How to find the acceleration of a ball rolling on ground?

[Eclair_de_XII]

Homework Statement

"A ball of moist clay falls 15.0 m to the ground. It is in contact with the ground for 20.0 ms before stopping. (a) What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground? (Treat the ball as a particle.) (b) Is the average acceleration up or down?"

Homework Equations

$x_0=15m$
$x=0m$
$t_2=0.02s$
$v^2=v_0+2a(x-x_0)$
$v=v_0+at$
Answer as given by book: (a) 857 m/s²; (b) up

The Attempt at a Solution

$v^2=2(9.8\frac{m}{s^2})(15m)$
$v_0=0\frac{m}{s}$
$v=17.15\frac{m}{s}$

Next, I treat this as my next initial velocity when the clay rolls on the ground for 0.02 s.

$v_0=17.15\frac{m}{s}$
$v=0\frac{m}{s}$
$t=0.02s$

$0\frac{m}{s}=|17.15\frac{m}{s}|+a(0.02s)$
$a=|\frac{-17.15\frac{m}{s}}{0.02s}|=|-875\frac{m}{s^2}|≠857\frac{m}{s^2}$

Aside from not matching the answer in the book, the sign of the acceleration is negative, which would mean it is going down. Unless I'm mistaken, and the negative acceleration is going up from -875 m\s² to 0 m\s²... I'm not sure if I'm correct or not, in this assumption, though. Could someone confirm or deny my answer and my assumption?

 

[haruspex]

When you solve v²= some number, what are the possible values of v?
I agree with the 17.15, as a magnitude. Check the next step. (Looks like you dropped a digit.)

 

[Eclair_de_XII]

Oh, I made a miscalculation.

$a=\frac{-17.15\frac{m}{s}}{0.02s}=-857.3\frac{m}{s^2}$
$|a|=857.3\frac{m}{s^2}$

So does that mean that the average acceleration is up, when it rises from -857 m/s² to 0 m/s²?

 

[haruspex]

Oh, I made a miscalculation.

$a=\frac{-17.15\frac{m}{s}}{0.02s}=-857.3\frac{m}{s^2}$
$|a|=857.3\frac{m}{s^2}$

So does that mean that the average acceleration is up, when it rises from -857 m/s² to 0 m/s²?

The acceleration does not rise from -857 m/s² to 0 m/s².
You did not answer my question about solving for v².

Edit: by the way, the ball does not roll on the ground, it squishes on the ground.

 

[Eclair_de_XII]

You did not answer my question about solving for v2.

It's going to be ±17.15 m/s. I also saw that I typed in 17.5 on my calculator; not 17.15.

 

[haruspex]

It's going to be ±17.15 m/s.

Right. Which is it here?

 

[Eclair_de_XII]

Negative, because it's going down?

 

[haruspex]

Negative, because it's going down?

Right. So what do you get for the acceleration?

 

[Eclair_de_XII]

-857 m/s²?

 

[haruspex]

-857 m/s²?

The velocity has gone from negative to zero. Does that make the acceleration positive or negative?

 

[Eclair_de_XII]

$a=\frac{v_f-v_i}{t}$
$a=\frac{0-(-17.15\frac{m}{s})}{0.02s}$
$a=857\frac{m}{s^2}$

So positive.

 

[Eclair_de_XII]

But that's only for the initial fall. What about after it touches the ground and squishes?

 

[haruspex]

But that's only for the initial fall. What about after it touches the ground and squishes?

No, you took the final velocity as zero to get that acceleration. That is the average acceleration while squishing.

 

[Eclair_de_XII]

Oh, so...

$v_0=-17.15\frac{m}{s}$
$v=0$
$t=0.02s$

$v=v_0+at$
$a=\frac{17.15\frac{m}{s}}{0.02s}=857\frac{m}{s^2}$

 

[haruspex]

Oh, so...

$v_0=-17.15\frac{m}{s}$
$v=0$
$t=0.02s$

$v=v_0+at$
$a=\frac{17.15\frac{m}{s}}{0.02s}=857\frac{m}{s^2}$

Yes.