How to find the acceleration of a rebounded ball?

[Eclair_de_XII]

Homework Statement

"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?"

Homework 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

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?

 

[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.

 

[Eclair_de_XII]

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

 

[haruspex]

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

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

 

[Eclair_de_XII]

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

 

[haruspex]

It goes to 2 m.

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

 

[Eclair_de_XII]

4.425 m/s.

 

[haruspex]

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

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

 

[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}##

 

[haruspex]

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}##

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).

 

[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.

 

[haruspex]

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.

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

 

[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}##

 

[haruspex]

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}##

What is the sign on the initial velocity!

 

[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}##

 

[haruspex]

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}##

Right, so what is the acceleration?

 

[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}##

 

[haruspex]

##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}##

Yes.