# How to find the acceleration of a rebounded ball?

1. Feb 26, 2016

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If the ball is in contact with the floor for 12.0 ms, (a) what is the magnitude of its average acceleration during that contact and (b) is the average acceleration up or down?"

2. Relevant equations
$x_0=4m$
$x=0m$
$a=9.8\frac{m}{s^2}$
$v^2=v_0^2+2a(x-x_0)$
$x-x_0=v_0t+\frac{1}{2}at^2$
Answer as given by book: (a) 1.26 ⋅ 103 m/s2; (b) up

3. The attempt at a solution
$v^2=2(9.8\frac{m}{s^2})(4m)=78.4\frac{m^2}{s^2}$
$v=8.85\frac{m}{s}$

Next, I change this final velocity for the fall from 4 m to 0 m to the initial velocity for the rebound from 0-m to 2 m.

$v_0=-8.85\frac{m}{s}$
$x=2m$
$x_0=0m$
$t=0.012s$

$2m=(8.85\frac{m}{s}(0.012s)+\frac{1}{2}a(0.012s)^2$
$4m-0.21251m=a(0.000144s^2)$
$a=26302.1\frac{m}{s^2}≠1260\frac{m}{s^2}$

Does something happen to the velocity of the ball once it touches the ground? Does it lose energy; if so, by how much?

2. Feb 26, 2016

### haruspex

The formula you applied for the second part of the calculation (a SUVAT formula) is for constant acceleration. But you applied it from the moment before the bounce to the top of the next bounce. Acceleration is certainly not constant over that period.
Find the velocity immediately after the bounce.

3. Feb 26, 2016

### Eclair_de_XII

Isn't it just 8.85 m/s? If not, that just brings me to that last question in my post.

4. Feb 26, 2016

### haruspex

How high does it go after bouncing? It does not reach 4m.

5. Feb 26, 2016

### Eclair_de_XII

It goes to 2 m. So what, is the velocity halved?

6. Feb 26, 2016

### haruspex

Right, so what is its speed immediately after the bounce?

7. Feb 26, 2016

### Eclair_de_XII

4.425 m/s.

8. Feb 26, 2016

### haruspex

You changed your post while I was writing mine...
No, the speed is not halved. Do the calculation.

9. Feb 26, 2016

### Eclair_de_XII

I really hate using the same equation too many times, but...

$x=2m$
$x_0=0m$
$v=0\frac{m}{s}$
$a=-9.8\frac{m}{s^2}$

$v^2=v_0^2+2(-9.8\frac{m}{s^2})(2m)$
$-v_0^2=(4m)(-9.8\frac{m}{s^2})$
$v_0^2=(4m)(9.8\frac{m}{s^2})$
$v_0=±6.621\frac{m}{s}$

10. Feb 26, 2016

### haruspex

Yes, but which sign is right?
You could have avoided the full computation by observing that, according to both the SUVAT equation and the KE+PE=constant equation, the height is proportional to the square of the speed, so if half the height just divide by sqrt(2).

11. Feb 26, 2016

### Eclair_de_XII

The positive one? I mean, since I already have 2 m as positive, in a direction going up, and the gravity as negative, going down, then it should be natural that the velocity of the ball, going up, should be positive.

12. Feb 26, 2016

### haruspex

Right, so what is the change in velocity caused by the bounce? Careful with signs.

13. Feb 26, 2016

### Eclair_de_XII

It's final minus initial velocity, right? So...

$Δv=v_f-v_i=(6.62\frac{m}{s}-8.85\frac{m}{s})=-2.23\frac{m}{s}$

14. Feb 26, 2016

### haruspex

What is the sign on the initial velocity!

15. Feb 26, 2016

### Eclair_de_XII

Oh, it's going down relative to 6.62 m/s, so it's negative.

$Δv=6.62\frac{m}{s}+8.85\frac{m}{s}=15.47\frac{m}{s}$

16. Feb 26, 2016

### haruspex

Right, so what is the acceleration?

17. Feb 26, 2016

### Eclair_de_XII

$a=\frac{15.47\frac{m}{s}}{0.012s}=1290\frac{m}{s^2}$

To be more precise...

$a=\frac{(78.4\frac{m^2}{s^2})^0.5+(39.2\frac{m^2}{s^2})^0.5}{0.012s}=1259.6\frac{m}{s^2}$

18. Feb 26, 2016

Yes.