Tennis Ball Dropped: Will It Ever Come to Rest?

[arpon]

Homework Statement

A tennis ball is dropped from a point 4.9 m above the ground. Every time it hits the
ground, it bounces back to ¾ of its previous height. How long will the ball take to
come to at rest? [ignore the time required for hitting the ground and turning back]

Homework Equations

The Attempt at a Solution

I think , it will never come to rest.

 

[Nathanael]

Interesting. I think that it will take an infinite amount of bounces to come to rest, but that does not necessarily mean that it will take an infinite amount of time. What do you think?

 

[haruspex]

I think , it will never come to rest.

Justify your answer with a calculation.

 

[TimeInquirer]

The thought of it always traveling 3/4ths the previous height should not make you think it will never be zero (at rest). Think of it as the height approaches 0.

 

[arpon]

Justify your answer with a calculation.

Interesting. I think that it will take an infinite amount of bounces to come to rest, but that does not necessarily mean that it will take an infinite amount of time. What do you think?

Yeah, I got it!
For the first drop, the required time is $\sqrt{ \frac {2h}{g}}$ ; [ as, $h = \frac{1}{2} gt^2$ ]
Then, the time interval between the first and second drop is $2 \sqrt{ \frac {2(\frac {3}{4} h)}{g}}$ ; [the time for going up to the maximum height of $\frac{3}{4} h$ and returning to ground]
In the same way, the time interval between the second and third drop is $2 \sqrt{ \frac {2((\frac {3}{4})^2 h)}{g}}$ , and so on.
Hence the total time required for coming to at rest is,
$t = \sqrt{ \frac {2h}{g}} + 2 \sqrt{ \frac {2(\frac {3}{4} h)}{g}} + 2 \sqrt{ \frac {2((\frac {3}{4})^2 h)}{g}} + ...$
And I can easily get the summation of this infinite geometric series, as the ratio is less than 1. But at the first sight at the problem, it seemed impossible for the ball to come to at rest because infinite number of drops are needed before coming to at rest. Would you please help me to visualize what actually happens?

 

[haruspex]

Yeah, I got it!
For the first drop, the required time is $\sqrt{ \frac {2h}{g}}$ ; [ as, $h = \frac{1}{2} gt^2$ ]
Then, the time interval between the first and second drop is $2 \sqrt{ \frac {2(\frac {3}{4} h)}{g}}$ ; [the time for going up to the maximum height of $\frac{3}{4} h$ and returning to ground]
In the same way, the time interval between the second and third drop is $2 \sqrt{ \frac {2((\frac {3}{4})^2 h)}{g}}$ , and so on.
Hence the total time required for coming to at rest is,
$t = \sqrt{ \frac {2h}{g}} + 2 \sqrt{ \frac {2(\frac {3}{4} h)}{g}} + 2 \sqrt{ \frac {2((\frac {3}{4})^2 h)}{g}} + ...$
And I can easily get the summation of this infinite geometric series, as the ratio is less than 1. But at the first sight at the problem, it seemed impossible for the ball to come to at rest because infinite number of drops are needed before coming to at rest. Would you please help me to visualize what actually happens?

Have you heard of Zeno's Achilles and the Tortoise paradox? Look it up.

 

[Nathanael]

$t = \sqrt{ \frac {2h}{g}} + 2 \sqrt{ \frac {2(\frac {3}{4} h)}{g}} + 2 \sqrt{ \frac {2((\frac {3}{4})^2 h)}{g}} + ...$

I would factor something out to make that less ugly!

Would you please help me to visualize what actually happens?

If you have a hard floor and some kind of bouncy ball, drop it on the floor. Listen to the way it bounces:
thud......thud......thud...thud...thud...thud...thud..thudthudthuthththttttt

 

[arpon]

I would factor something out to make that less ugly!

I did so in my exercise book, but here, I just copied and pasted the terms I typed previously.;)

If you have a hard floor and some kind of bouncy ball, drop it on the floor. Listen to the way it bounces:
thud......thud......thud...thud...thud...thud...thud..thudthudthuthththttttt

But, it seems awkward that in the last second there is infinite number of drops.

 

[Nathanael]

But, it seems awkward that in the last second there is infinite number of drops.

Well in the real world this doesn't happen of course. Only in a perfectly ideal world where the bounces are somehow always 3/4 the previous height.
But yes I agree, even in an ideal world, it seems a bit strange. I would take up haruspex on his advice and look up that paradox (or one of it's many variations).

 

[Nathanael]

But, it seems awkward that in the last second there is infinite number of drops.

The key point (which I forgot to mention in my last post) is that the problem said to neglect the time between bounces. If the time between bounces was finite (no matter how small!) then it would take an infinite amount of time.

This is how I justify it to my self as making "sense" :)

 

[arpon]

The key point (which I forgot to mention in my last post) is that the problem said to neglect the time between bounces. If the time between bounces was finite (no matter how small!) then it would take an infinite amount of time.

This is how I justify it to my self as making "sense" :)

I agree with you.:)

 

[haruspex]

The key point (which I forgot to mention in my last post) is that the problem said to neglect the time between bounces. If the time between bounces was finite (no matter how small!) then it would take an infinite amount of time.

Couple of points...
I would word that as 'the time taken to bounce', and I don't think you mean 'finite', or even 'nonzero'. The time taken to bounce could be reducing geometrically in the same way that the times in the air are.