- #1

- 12

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

In summary, the conversation discusses graphing the heights of a bouncing rubber ball on the y-axis and the time taken for it to bounce on the x-axis. The function is quadratic, but the reason for this is not known. It is suggested that the displacement is related to time and the ball's motion has constant acceleration. The concept of energy loss is introduced and it is explained that the leading term of the function will still be quadratic. The conversation also briefly mentions making the function linear and the coefficient of restitution.

- #1

- 12

- 0

Physics news on Phys.org

- #2

- 40

- 6

What do you exactly mean by the time taken for it to bounce? Is it the time taken for it to reach to the ground or to reach the ground and bounce a few times and stop due to loss of energy?Jeven said:

- #3

Homework Helper

- 15,542

- 1,916

It is motion with constant acceleration. How is the displacement related to time?Jeven said:

- #4

- 12

- 0

I am sorry I forgot to add that the time is theGuneykan Ozgul said:What do you exactly mean by the time taken for it to bounce? Is it the time taken for it to reach to the ground or to reach the ground and bounce a few times and stop due to loss of energy?

- #5

- 40

- 6

Okay. Now, let's say that acceleration is constant a . Now the velocity at time t will be at. Then the displacement d=∫atdt=1/2at^2 assuming that the initial position is 0.Jeven said:I am sorry I forgot to add that the time is thetime taken for it to bounce 5 times.

Now let's say that the ball hits the ground and the energy of the ball reduces to bE where E is the energy of the ball before it hits the ground and b is a some arbitrary constant. Now the particle will go to 1/b of initial height then the time will be √(1/b) of the time it takes to hit the ground. So if you do this 5 times you will find the total time. If b is 1, that is ball does not lose energy when it hits the ground you will get directly get only quadratic term since d is proportional to square of t. If b is bigger than 1, still the leading term will be square of t. So it is a quadratic.

Hope this is helpful.

- #6

- 12

- 0

Yes it is thank you. But I don't quite understand the (1/b) part, why would that be the energy? And what would I do to make this function a linear function? Because I need to graph a straight line.Guneykan Ozgul said:Okay. Now, let's say that acceleration is constant a . Now the velocity at time t will be at. Then the displacement d=∫atdt=1/2at^2 assuming that the initial position is 0.

Now let's say that the ball hits the ground and the energy of the ball reduces to bE where E is the energy of the ball before it hits the ground and b is a some arbitrary constant. Now the particle will go to 1/b of initial height then the time will be √(1/b) of the time it takes to hit the ground. So if you do this 5 times you will find the total time. If b is 1, that is ball does not lose energy when it hits the ground you will get directly get only quadratic term since d is proportional to square of t. If b is bigger than 1, still the leading term will be square of t. So it is a quadratic.

Hope this is helpful.

- #7

- 40

- 6

- #8

- 1

- 0

Guneykan Ozgul said:Okay. Now, let's say that acceleration is constant a . Now the velocity at time t will be at. Then the displacement d=∫atdt=1/2at^2 assuming that the initial position is 0.

Now let's say that the ball hits the ground and the energy of the ball reduces to bE where E is the energy of the ball before it hits the ground and b is a some arbitrary constant. Now the particle will go to 1/b of initial height then the time will be √(1/b) of the time it takes to hit the ground. So if you do this 5 times you will find the total time. If b is 1, that is ball does not lose energy when it hits the ground you will get directly get only quadratic term since d is proportional to square of t. If b is bigger than 1, still the leading term will be square of t. So it is a quadratic.

Hope this is helpful.

if b is the coefficient of restitution, COR, which is less than 1,then why would it go to 1/b of the initial height, meaning it would surpass the initial height since 1/COR>1 ?? I see you know your stuff so I must be at a loss here so could you please explain your meaning :)

- #9

- 51

- 12

Belovedcritic said:if b is the coefficient of restitution, COR, which is less than 1,then why would it go to 1/b of the initial height, meaning it would surpass the initial height since 1/COR>1 ?? I see you know your stuff so I must be at a loss here so could you please explain your meaning :)

He just made some little confusion I guess.

If the energy of the ball at the initial peak [itex]h_0[/itex] is [itex]E_0 = mgh_0[/itex], after the bounce you'll have, again at the peak, [itex]E_1 = bE_0 = mgbh_0[/itex], so [itex]h_1=bh_0[/itex].

The function for the velocity of a bouncing ball is a mathematical equation that describes the relationship between the velocity of the ball and the number of bounces it makes. It is typically represented as v = √(gh) where v is the velocity, g is the acceleration due to gravity, and h is the height of the ball.

The function for the velocity of a bouncing ball is derived from the principles of conservation of energy and Newton's laws of motion. It takes into account the initial velocity of the ball, the height from which it is dropped, and the effects of gravity and air resistance.

The factors that can affect the velocity function of a bouncing ball include the initial velocity, the height from which it is dropped, the surface on which it bounces, and the presence of air resistance. These factors can alter the amount of energy the ball has and how it is transferred during each bounce.

The velocity of a bouncing ball decreases with each bounce due to the conversion of kinetic energy to potential energy and back. As the ball bounces, it loses energy to air resistance and heat, resulting in a lower velocity with each subsequent bounce.

Yes, the function for the velocity of a bouncing ball can be used to predict its future bounces, assuming that the environmental factors remain constant. By plugging in different values for the initial velocity and height, we can calculate the velocity of the ball at any given point during its bounces.

Share:

- Replies
- 19

- Views
- 1K

- Replies
- 3

- Views
- 557

- Replies
- 3

- Views
- 542

- Replies
- 7

- Views
- 469

- Replies
- 13

- Views
- 1K

- Replies
- 5

- Views
- 581

- Replies
- 10

- Views
- 1K

- Replies
- 3

- Views
- 557

- Replies
- 14

- Views
- 1K

- Replies
- 4

- Views
- 531