# Function for the velocity of a bouncing ball

I graphed different heights from which I had dropped a bouncing rubber ball on the y-axis and the time taken for it to bounce on the x-axis. The function came out to be quadratic, but I do not know why. If someone can show mathematically why this is, that'd be splendid. Thank you.

I graphed different heights from which I had dropped a bouncing rubber ball on the y-axis and the time taken for it to bounce on the x-axis. The function came out to be quadratic, but I do not know why. If someone can show mathematically why this is, that'd be splendid. Thank you.
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?

ehild
Homework Helper
I graphed different heights from which I had dropped a bouncing rubber ball on the y-axis and the time taken for it to bounce on the x-axis. The function came out to be quadratic, but I do not know why. If someone can show mathematically why this is, that'd be splendid. Thank you.
It is motion with constant acceleration. How is the displacement related to time?

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?
I am sorry I forgot to add that the time is the time taken for it to bounce 5 times.

I am sorry I forgot to add that the time is the time taken for it to bounce 5 times.
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.

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.
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.

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