Tennis Ball Dropped: Will It Ever Come to Rest?

In summary: In the last second, the ball will have gone up 3/4ths of the way, turned around, and come back down. So in reality, the ball will have gone up and down an infinite number of times.
  • #1
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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.
 
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  • #2
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?
 
  • #3
arpon said:
I think , it will never come to rest.
Justify your answer with a calculation.
 
  • #4
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.
 
  • #5
haruspex said:
Justify your answer with a calculation.
Nathanael said:
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?
 
  • #6
arpon said:
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.
 
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  • #7
arpon said:
##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!

arpon said:
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
 
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  • #8
Nathanael said:
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.;)
Nathanael said:
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. o_O
 
  • #9
arpon said:
But, it seems awkward that in the last second there is infinite number of drops. o_O
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).
 
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  • #10
arpon said:
But, it seems awkward that in the last second there is infinite number of drops. o_O
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" :)
 
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  • #11
Nathanael said:
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.:)
 
  • #12
Nathanael said:
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.
 
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1. Will a tennis ball ever stop bouncing?

Yes, eventually a tennis ball will come to rest and stop bouncing. The time it takes for a tennis ball to stop bouncing depends on various factors, such as the height from which it was dropped, the type of surface it lands on, and the air resistance.

2. Why does a tennis ball keep bouncing?

A tennis ball keeps bouncing because of its elasticity. When the ball hits the ground, it compresses and stores potential energy. As it bounces back up, this potential energy is converted into kinetic energy, causing the ball to bounce back up. This process continues until all the energy is dissipated and the ball comes to rest.

3. Can a tennis ball bounce forever?

No, a tennis ball cannot bounce forever. Due to the law of conservation of energy, the energy of the ball decreases with each bounce as it is converted into other forms, such as heat and sound. Eventually, the ball will lose all its energy and come to rest.

4. Does the height from which a tennis ball is dropped affect its bounce?

Yes, the height from which a tennis ball is dropped does affect its bounce. The higher the ball is dropped from, the more potential energy it has, and the higher it will bounce. This is because the ball has more distance to accelerate towards the ground, resulting in a greater impact force.

5. Why does a tennis ball bounce higher on some surfaces than others?

A tennis ball bounces higher on some surfaces than others due to the surface's elasticity. Softer and more elastic surfaces, such as grass or rubber, will absorb less energy from the ball, allowing it to bounce higher. Harder surfaces, like concrete or wood, are less elastic and absorb more energy, resulting in a lower bounce.

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