Quadratic Formula Homework: Meaning of A, B & C in Y = At^2 + Bt + C

[mlostrac]

Homework Statement

What are the meanings of each of the parameters in the Y = At^2 + Bt + C equation?

example:
A = -4.8
B = 21.9
C = -23.8

I know that B is suppose to represent the initial velocity, but not sure about the other 2. Is A supposed to represent acceleration?

 

[diazona]

Are you thinking of this?
$$y = y_0 + v_0 t + \frac{1}{2}at^2$$
In that equation v₀ is indeed initial velocity, y₀ is initial position, and a is acceleration.

 

[mlostrac]

Are you thinking of this?
$$y = y_0 + v_0 t + \frac{1}{2}at^2$$
In that equation v₀ is indeed initial velocity, y₀ is initial position, and a is acceleration.

No, I am thinking of the original equation I posted. Are the two somehow related? The textbook I have says that B is interpreted as the initial velocity of the object who's height versus time graph forms an inverted U parabola

 

[rock.freak667]

No, I am thinking of the original equation I posted. Are the two somehow related? The textbook I have says that B is interpreted as the initial velocity of the object who's height versus time graph forms an inverted U parabola

Well don't the two look at least similar?

Also you can interpret what the coefficients are by the units.

You know that the left side's units are m. Therefore, all the other terms must the sum of quantities with units of m.

So for 'Bt', t has units seconds, s.

Bt must have units m.

The only way for 'Bt' to have units as 'm' is for B to have m/s as the units. Understand?

 

[mlostrac]

Well don't the two look at least similar?

Also you can interpret what the coefficients are by the units.

You know that the left side's units are m. Therefore, all the other terms must the sum of quantities with units of m.

So for 'Bt', t has units seconds, s.

Bt must have units m.

The only way for 'Bt' to have units as 'm' is for B to have m/s as the units. Understand?

Ok, so A must represent acceleration since its At^2

And I'm going to guess and say that the C parameter is just a y-value representing height since it's only in meters. So I'm guessing that's the height of the inverted U parabola?

 

[RTW69]

Look closer at diazona's post. C is Yo, the initial position which is 23.8 meters below the origin.

 

[clombard1973]

Homework Statement

What are the meanings of each of the parameters in the Y = At^2 + Bt + C equation?

example:
A = -4.8
B = 21.9
C = -23.8

I know that B is suppose to represent the initial velocity, but not sure about the other 2. Is A supposed to represent acceleration?

Hi there,
Yeah Diazona is correct. If you are stating that the initial velocity is B, then your equation is the equation for distance given an acceleration upon the mass, an initial velocity of the mass and some initial height of the mass.

It is a very common equation used for finding the distance an object may obtain given it is undergoing an acceleration and/or has an initial velocity. If it has no acceleration acting on it nor an initial velocity , both A and B are zero which means the position of your mass is just the initial position, C.

If you do not know the answer and someone gives it to you in a form that is even more explicit than the one you have, but is obviously the same function, you shouldn't state it to be incorrect unless you know it is; and it is not incorrect.

A is not the acceleration, it is twice the acceleration, because it is supposed to be 1/2at^2 and A = 1/2a where a is the acceleration acting upon the mass, so A is twice the acceleration. Bt is the same as Vot, the initial velocity times the time, so B is the initial velocity and C is just the initial position of the mass before it has moved, or the position at time equals zero if you wish, but before you tell someone they are incorrect you should be sure they are first; because in this case Diazona was correct and gave you the formula in the more explicit form, as opposed to simply the constant terms of A, B and C that you are using. You asked what they were and that person told you, and you said no I mean my equation; your equation and his/hers are the same, and in fact the equation that Diazona gave to you tells you what A, B and C are.

Look a little more carefully next time.

Craig

 

[mlostrac]

Hi there,
Yeah Diazona is correct. If you are stating that the initial velocity is B, then your equation is the equation for distance given an acceleration upon the mass, an initial velocity of the mass and some initial height of the mass.

It is a very common equation used for finding the distance an object may obtain given it is undergoing an acceleration and/or has an initial velocity. If it has no acceleration acting on it nor an initial velocity , both A and B are zero which means the position of your mass is just the initial position, C.

If you do not know the answer and someone gives it to you in a form that is even more explicit than the one you have, but is obviously the same function, you shouldn't state it to be incorrect unless you know it is; and it is not incorrect.

A is not the acceleration, it is twice the acceleration, because it is supposed to be 1/2at^2 and A = 1/2a where a is the acceleration acting upon the mass, so A is twice the acceleration. Bt is the same as Vot, the initial velocity times the time, so B is the initial velocity and C is just the initial position of the mass before it has moved, or the position at time equals zero if you wish, but before you tell someone they are incorrect you should be sure they are first; because in this case Diazona was correct and gave you the formula in the more explicit form, as opposed to simply the constant terms of A, B and C that you are using. You asked what they were and that person told you, and you said no I mean my equation; your equation and his/hers are the same, and in fact the equation that Diazona gave to you tells you what A, B and C are.

Look a little more carefully next time.

Craig

It's looks like you're saying that I was rude in my answer to diazona, which I did not intend to be. I was simply replying because I actually did not know that the 2 equations would be related in this case.

I was confused with what is being asked in my course and still am not 100% sure on the situation. B and C make sense. And it makes sense that the quadratic formula and the kinematic equation are related. But...

Say a ball is tossed straight up in the air and then caught, hence making a parabola. The quadratic formula has an A value of -4.8. I know the negative determines the shape of the parabola (hump face up), but all I want to know is if A represents accleration.

At^2, with A=-4.8, whereas in the kinematic equation 1/2At^2

BUT

4.8t^2 is the same as 1/2(9.8)t^2, is it not? So I don't know what u mean when you say it's twice the acceleration because they are equal?

 

[diazona]

No offense taken, by the way

Let me see if I'm understanding you correctly. You're saying that you have an equation which specifies the height, y, of a ball as a function of time, t. In that equation, you have a term which is a constant (number) times t², another term which is a different constant times t, and a third term which is just a constant, without any t. You've decided to label the first constant A, the second constant B, and the third constant C. In effect, you're saying that you have an equation of the form
$$y = At^2 + Bt + C$$
for certain numeric values of A, B, and C. Am I correct so far?Assuming that's all good, take a look at the equation I posted,
$$y = y_0 + v_0 t + \frac{1}{2}at^2$$
This is the equation that physicists use to describe objects moving with constant acceleration. That is precisely the situation you're describing, so this equation should apply.As you've figured out, $y_0$ in that equation is equivalent to your C. How did you (or could you) figure that out? Well, you see that in each equation, there is exactly one constant term (i.e. one term which doesn't have a t in it). The two equations must be the same, since they describe the same physical situation, and if the equations are the same, the constant terms must be the same. In your notation, you label that constant C, and in my notation, I label it $y_0$, but it's the same number underneath the label. So you can say that $C = y_0$.The same applies for the linear term, the one with a constant times t. In your notation, that constant is B, and in my notation, that constant is $v_0$, but again, it should be the same number underneath the label. That tells you that $B = v_0$.Now look at the quadratic term, the one with a constant times t². You can apply the same reasoning again. In your notation, that constant is labeled A, but in my notation, it's labeled $\frac{1}{2}a$ (note that uppercase A and lowercase a are not the same). So can you write the equation that relates your A to my $\frac{1}{2}a$? (By the way, all this will make more sense once you think through it, as opposed to me just telling you)Finally, I will tell you that lowercase a represents acceleration. Knowing that, what can you conclude about how your constant A is related to the acceleration? If A=-4.8m/s², what is the acceleration?

 

[clombard1973]

Well said

Yup I like it. Very descriptive in your reply, you relate each of his constants to yours and state the meaning of yours, i.e. initial position, initial velocity, gravitational acceleration on a mass here due to Earth.

There really isn't anything I can add to that last reply, other than telling you what the acceleration is, but that I think you can figure out. Just remember, the function describes the position of the ball versus time, so if you were to graph that you would get an inverted parabola as the ball leaves your hand, hits some maximum height, and then accelerates down to Earth (Earth is accelerating on it at all times but initially the ball has the momentum to overcome the downward pull from gravity and keep moving upwards for some time until gravity wins and pulls the ball back down).

Craig

 

[mlostrac]

No offense taken, by the way

Let me see if I'm understanding you correctly. You're saying that you have an equation which specifies the height, y, of a ball as a function of time, t. In that equation, you have a term which is a constant (number) times t², another term which is a different constant times t, and a third term which is just a constant, without any t. You've decided to label the first constant A, the second constant B, and the third constant C. In effect, you're saying that you have an equation of the form
$$y = At^2 + Bt + C$$
for certain numeric values of A, B, and C. Am I correct so far?


Assuming that's all good, take a look at the equation I posted,
$$y = y_0 + v_0 t + \frac{1}{2}at^2$$
This is the equation that physicists use to describe objects moving with constant acceleration. That is precisely the situation you're describing, so this equation should apply.


As you've figured out, $y_0$ in that equation is equivalent to your C. How did you (or could you) figure that out? Well, you see that in each equation, there is exactly one constant term (i.e. one term which doesn't have a t in it). The two equations must be the same, since they describe the same physical situation, and if the equations are the same, the constant terms must be the same. In your notation, you label that constant C, and in my notation, I label it $y_0$, but it's the same number underneath the label. So you can say that $C = y_0$.


The same applies for the linear term, the one with a constant times t. In your notation, that constant is B, and in my notation, that constant is $v_0$, but again, it should be the same number underneath the label. That tells you that $B = v_0$.


Now look at the quadratic term, the one with a constant times t². You can apply the same reasoning again. In your notation, that constant is labeled A, but in my notation, it's labeled $\frac{1}{2}a$ (note that uppercase A and lowercase a are not the same). So can you write the equation that relates your A to my $\frac{1}{2}a$? (By the way, all this will make more sense once you think through it, as opposed to me just telling you)


Finally, I will tell you that lowercase a represents acceleration. Knowing that, what can you conclude about how your constant A is related to the acceleration? If A=-4.8m/s², what is the acceleration?

I would say that the accleration is 9.8 m/s^2 because it's explaining an object in free fall.

If I have a bouncing ball, and each bounce is represented using the quadratic formula, does it make sense that the C value can be a high negative number (like the numbers I posted in my very first post)?

The ball only traveled a meter high, yet my C value (or Yo value) is -23.8? I was assuming that was in meters, but that really doesn't make sense

 

[clombard1973]

There's nothing wrong with a very small initial position

I would say that the accleration is 9.8 m/s^2 because it's explaining an object in free fall.

If I have a bouncing ball, and each bounce is represented using the quadratic formula, does it make sense that the C value can be a high negative number (like the numbers I posted in my very first post)?

The ball only traveled a meter high, yet my C value (or Yo value) is -23.8? I was assuming that was in meters, but that really doesn't make sense

If the initial position of the ball at time t = 0 is -23.8 meters, that just means with respect to the place where the measurement of position is being taken is 23.8 meters above the spot where the ball is presumed to be when the experiment begins and time starts moving.

So there's nothing wrong with having an initial position of -23.8 meters; it is completely relative to where you are determining the position of the ball. For instance if you were on a step ladder about 23.8 meters above the ground and the ball was shot out of an air cannon with an initial velocity of 21.9 meters/second at time t = 0, then to you the initial position of the ball with respect to where you are measuring its position is -23.8 meters below you and the ball shoots up with its initial velocity getting it to some height before gravity wins out and starts "pulling" the ball down wards. It is always pulling the ball down wards, but initially it has enough velocity in the upward direction so that gravity must first overcome that initial velocity, slow it to a stop, and then it will fall back towards Earth.


Having an initial position of -23.8 meters just means that the measurement is with respect to a place that happens to be 23.8 meters above where the ball is initially "shot" out with some initial velocity.

So if you got a maximum height from your equation of 1 meter, that is not how high it traveled, it traveled that plus the 23.8 meters it was set below you before the experiment began. You can take the equation, take the derivative of it with respect to time and set it equal to zero, find the time when the ball hits its maximum height and place that time back into the original function. If it gives you 1 meter for an answer, remember the ball first had to travel 23.8 meters to get to a height equal to zero, with respect to where you are measuring it from, and then it goes another meter above that meaning the ball traveled a total height of 24.8 meters before stopping and falling back to the ground.

If it has an initial velocity (assuming what was given was the vertical initial velocity) of 21.9 meters/sec and Earth is changing the velocity of the ball by 9.8 meters per second every second, then in one seconds time the ball wants to travel a full 21.9 meters high, but gravity slows it some over that seconds worth of time such that after 1 second instead of reaching a height of 21.9 meters, it may only get to about 17 meters. Every second after that it slows by another 9.8 meters per second, until eventually it no longer even has an upwards speed at which point the ball starts falling back towards the ground.

There's no doubt with an initial velocity of 21.9 meters per second that is upwards that it will travel further than 1 meter before falling to Earth. My guess is you found the maximum height the ball reached but forgot that it was already 23.8 meters below you before it went 1 meter above you.

Many Smiles,
Craig

 

[mlostrac]

If the initial position of the ball at time t = 0 is -23.8 meters, that just means with respect to the place where the measurement of position is being taken is 23.8 meters above the spot where the ball is presumed to be when the experiment begins and time starts moving.

So there's nothing wrong with having an initial position of -23.8 meters; it is completely relative to where you are determining the position of the ball. For instance if you were on a step ladder about 23.8 meters above the ground and the ball was shot out of an air cannon with an initial velocity of 21.9 meters/second at time t = 0, then to you the initial position of the ball with respect to where you are measuring its position is -23.8 meters below you and the ball shoots up with its initial velocity getting it to some height before gravity wins out and starts "pulling" the ball down wards. It is always pulling the ball down wards, but initially it has enough velocity in the upward direction so that gravity must first overcome that initial velocity, slow it to a stop, and then it will fall back towards Earth.


Having an initial position of -23.8 meters just means that the measurement is with respect to a place that happens to be 23.8 meters above where the ball is initially "shot" out with some initial velocity.

So if you got a maximum height from your equation of 1 meter, that is not how high it traveled, it traveled that plus the 23.8 meters it was set below you before the experiment began. You can take the equation, take the derivative of it with respect to time and set it equal to zero, find the time when the ball hits its maximum height and place that time back into the original function. If it gives you 1 meter for an answer, remember the ball first had to travel 23.8 meters to get to a height equal to zero, with respect to where you are measuring it from, and then it goes another meter above that meaning the ball traveled a total height of 24.8 meters before stopping and falling back to the ground.

If it has an initial velocity (assuming what was given was the vertical initial velocity) of 21.9 meters/sec and Earth is changing the velocity of the ball by 9.8 meters per second every second, then in one seconds time the ball wants to travel a full 21.9 meters high, but gravity slows it some over that seconds worth of time such that after 1 second instead of reaching a height of 21.9 meters, it may only get to about 17 meters. Every second after that it slows by another 9.8 meters per second, until eventually it no longer even has an upwards speed at which point the ball starts falling back towards the ground.

There's no doubt with an initial velocity of 21.9 meters per second that is upwards that it will travel further than 1 meter before falling to Earth. My guess is you found the maximum height the ball reached but forgot that it was already 23.8 meters below you before it went 1 meter above you.

Many Smiles,
Craig

K well I see what you mean with your example. But I dropped the ball from chest height (with a motion sensor facing down on it). All of the measurements it the computer program took were between 0 and 1.3 meters. So it's kind of hard to imagine and relate the numbers to the situation. So that's why I still find it a bit odd and confusing.

Or is the Yo value related to the Vo value. Because if you look at my initial values, B and C are almost the same ( 21.9 and -23.8 respectively).

 

[diazona]

Think about what units those values have. The fact that the numbers are close together is meaningless by itself, if you don't also match the units.

 

[clombard1973]

K well I see what you mean with your example. But I dropped the ball from chest height (with a motion sensor facing down on it). All of the measurements it the computer program took were between 0 and 1.3 meters. So it's kind of hard to imagine and relate the numbers to the situation. So that's why I still find it a bit odd and confusing.

Or is the Yo value related to the Vo value. Because if you look at my initial values, B and C are almost the same ( 21.9 and -23.8 respectively).

The fact that B and C are almost the same, whatever units are used, is at best a coincidence. They have nothing to do with each other and are NOT related. Nonetheless diazano is correct, something is not right because this equation has been used for almost 300 years to accurately describe the position of a mass under the influence of gravitational acceleration (any acceleration for that matter) with an initial velocity and position.

How can you be dropping a ball whose initial velocity is positive? Gravity pulls down on it and the A term is -0.5*9.8m/s^2, it is negative because the force is acting to pull the ball down wards. A C term of -23.8 implies the ball must start from a position that is 23.8 meters below the point you are taking measurements from. You have numbers for something that is shot upwards at an initial speed of 21.9 m/s with an initial height of -23.8 meters, but then you say you are dropping the ball. That makes no sense.

Where did you get A, B and C? Think about it. If B is positive it means it starts at time t = 0 moving in the opposite direction that gravity is acting upon it. Gravity wants to pull it down, but it has an initial velocity that is upwards, so you cannot drop it and use that equation with those numbers. The numbers you have for A, B and C imply that something at time t = 0 is sent vertically upward with an initial velocity of 21.9 m/s starting from a height of y(0) = -23.8 and is constantly changing speed by 9.8m/s every second.

So the equation and numbers you have with these initial conditions of the experiment can predict the position of the ball at any time, t. The initial conditions you have is that it is moving at a speed of 21.9 something/second, I say something because it could be in feet per second, meters per second, yards per second, that is up to you as to what units you wish to use to measure these things, but given that A = -4.8 and g in meters/sec^2 is -9.8, probably means that you are using stuff in meters and in seconds. So if that is the case and you are dropping a ball from some height then what's the initial velocity? When you let go of the ball from some height how fast is it moving the moment you release it?

ZERO m/s. It is not moving the moment you let go of it so it has no initial velocity, so B would be zero and not 21.9. You need to check what it is you are trying to describe and apply the equation correctly. If you are simply dropping the ball then there is no initial velocity and the initial height you can choose as either zero, or what it is above the ground, the function will tell you the position of the ball as it is moving at any time, t, with respect to where you define the height to be zero.

You need to first understand what it is you are even looking for. You give a very well known physics equation along with the constants in front and then when you are told what those constants represent, it's almost as if you are arguing with the facts of the equation. The equation is never wrong, how you apply it may well be. A represents 1/2 times the acceleration due to gravity. B is the initial velocity. C is the initial position. If any of the values you gave do not match up with what you are using during the measurement part of this experiment, then the equation cannot accurately predict the position of the ball at any time, t.

Is the ball starting with an initial height of -23.8 whatever, e.g. feet, meters, yards, angstroms (that one was a joke). If it is not, do not use it to predict the position of the ball. Is the ball moving at time t = 0 upwards with a speed of 21.9 whatever/second? If it is not do not use that number as the initial velocity.

You need to setup the initial function such that C is the initial position of the ball, B is the initial velocity of the ball (and if it is positive that means at time t = 0 the ball is moving in the opposite direction gravity is acting upon it, and gravity wants to pull it down, so an initial velocity that is positive implies it is first moving upwards before it stops and descends back down).

A is 1/2g, now you can use g as -32 ft/sec^2 or -9.8 m/sec^2 or whatever units you'd like as long as every term in the function has the same units so they can sum to a final position given a time value. If you are going to use g as -9.8 m/s^2 then everything else must be in meters and seconds. So B is an initial velocity in meters per second. Did the ball shoot upwards at time t = 0 with an initial speed of 21.9 m/s? If not you are using the wrong number for B. Where was the ball at time t = 0? If it was not 23.8 meters below where you were taking measurements from then you are using the wrong value of C.

I cannot answer a question when you keep replying with things like, but I dropped the ball from chest height. That tells me that it had no initial velocity and was not at an initial height of -23.8, but you are giving numbers for B and C that both imply an initial upwards velocity and an initial position of the ball that is -23.8 units below where you are taking measurements from.

You can't just grab a function and hope it works for every case, you need to find the initial values needed for the function to accurately describe the position of the ball at any time, t. You need to put in the initial velocity and initial position. It CANNOT have an initial velocity that is upwards if you are dropping it from chest height. Think about what you are trying to do. You wish to use a function to describe the position of a ball given an initial set of conditions. That function will do the job if it is given the correct initial values.

Either you are not doing the experiment according to how the initial setup is supposed to be; that is with the ball initially -23.8 units below where you are measuring from that starts with an initial speed of 21.9 units/sec upwards and is being pulled down wards the entire time by -4.8 units/sec^2, or you are using the incorrect constants out in front. It doesn't matter if your professor gave you those values or not. If you are not performing the experiment according to those initial conditions you will not be able to accurately describe the position of the ball with that function, performing the experiment the way you are. You need to set up the experiment such that the ball is -23.8 units below where you are measuring from that has an initial velocity of 21.9 units/sec upwards and is being pulled down by gravity at a rate of -4.5units/sec^2.

If your setup is not this way when you begin, the function you are using is not worth the paper it is written on. It is important when performing an experiment that you wish to predict the outcome of before hand (that is what science is all about), that you first setup the function with the correct initial values if you wish that function to be of any use whatsoever.

If you are simply dropping the ball from some height, then you are not using the correct initial conditions and the function will give you answers that are worse than simply taking guesses at what the position or velocity of the ball may be at any moment in time.

If you are looking for an answer, then you need to tell us what is the initial position of the ball. What is the initial velocity of the ball. You gave us numbers for B and C that would tell us that and when I tell you what the initial velocity and position must be according to the numbers you provided you say it can't be because you dropped it from chest height. Well if you dropped it from chest height how can it have such a large upwards initial velocity?

I do not think you know how to relate the initial conditions within the function to that which you are using during the measuring portion of the experiment. Please make up your mind. Does it have an initial velocity that is upwards at 21.9 units/sec? Where does the ball sit at time t = 0? You need those values for B and C inside your function for it to be of any use at all.

Craig

 

[mlostrac]

"I cannot answer a question when you keep replying with things like, but I dropped the ball from chest height. That tells me that it had no initial velocity and was not at an initial height of -23.8, but you are giving numbers for B and C that both imply an initial upwards velocity and an initial position of the ball that is -23.8 units below where you are taking measurements from."

No, I don't think you get what I'm saying. When I said I dropped it from chest height, I meant that the ball never reached any extreme heights (-23), it just bounced between the ground to a height no taller than my body.

As for the 3 values (A B C), they were generated with the parabola that looked like this http://www.devmaster.net/articles/fast-sine-cosine/parabola.gif (ignore the units).

The beginning of the parabola is when the ball (after being dropped by myself) bounced off the ground, to it's max height (the peak of the parabola) and then back down to the ground.

After it bounced a second time, another parabola was generated representing from when the bounce occurred to when it hit the ground for a third time.

So the values represent the ball after a bounce until the next bounce.

Do you know what I mean?

 

[vela]

You're going to have to go back to your data and analysis and work out the units of A, B, and C. While A seems to be in m/s², the other two seem off in light of the physical situation you described if you are indeed using meters and seconds.

 

[clombard1973]

"I cannot answer a question when you keep replying with things like, but I dropped the ball from chest height. That tells me that it had no initial velocity and was not at an initial height of -23.8, but you are giving numbers for B and C that both imply an initial upwards velocity and an initial position of the ball that is -23.8 units below where you are taking measurements from."

No, I don't think you get what I'm saying. When I said I dropped it from chest height, I meant that the ball never reached any extreme heights (-23), it just bounced between the ground to a height no taller than my body.

As for the 3 values (A B C), they were generated with the parabola that looked like this http://www.devmaster.net/articles/fast-sine-cosine/parabola.gif (ignore the units).

The beginning of the parabola is when the ball (after being dropped by myself) bounced off the ground, to it's max height (the peak of the parabola) and then back down to the ground.

After it bounced a second time, another parabola was generated representing from when the bounce occurred to when it hit the ground for a third time.

So the values represent the ball after a bounce until the next bounce.

Do you know what I mean?

It is almost as if you are not reading what I am telling you. The equation you are using has to be set up for the experiment you are doing. You say it has values representing the position of the ball after a bounce, when? 10 seconds after a bounce? 15 seconds after a bounce, or just after a bounce. And the equation is only good for something that is in under a constant acceleration, so it cannot describe the position of the ball after it strikes the ground without setting up the equation with new values once again because the "initial" velocity (the velocity just after the bounce) has changed from what it was when released and the initial position is no longer at chest height but is at ground level.

In other words you cannot use it to describe the height of the ball if it first hits the ground and bounces back, the equation does not know you are doing that. The equation thinks it is in free fall after overcoming perhaps an initial velocity upwards, or downwards.

Even if you generated your own constants in front how did you get those values? What do they represent to you; I have told you more than once what they are? A is supposed to 1/2 the acceleration the ball is under at all times, and it assumes that acceleration is constant; if the ball strikes the ground and bounces back it is not under a constant acceleration; it is before it hits the ground and afterwards, but not during. The B term is your initial velocity, you have 21.9 m/s, did the ball just after striking the ground bounce back up with a speed of 21.9 m/s? Probably not if you dropped it onto the floor.

The initial position of the ball just after it hits the floor you have as -23.8 m. So are you starting your experiment when the ball hits something that is 23.8 meters below the place from where you dropped the ball and the place you are measuring the position of the ball is therefore 23.8 meters above where the ball first strikes the ground and bounces back with some initial velocity; because that is how you have the equation setup using those values? I doubt it, you are not using the equation correctly. That is what I have been telling you but you keep responding defending your usage of the equation when it obviously is not giving you the correct answers, again why? BECAUSE YOU NEED TO USE IT PROPERLY!

I'll try this one last time and if you just do not get it after this, you'll have to ask someone else who may get through to you. The graph you have attached to your last reply, let's say it represents the position (the Y position) of the ball versus time (even though the horizontal axis is labeled with x, you want the maximum height the ball reaches, so let's work with only the vertical velocities). It clearly starts off with an initial position of zero, so C should be zero, not -23.8 like you have been saying; if you are using that graph as your "template" which the ball follows versus time and is what you want to find and describe at any point in time just after the first bounce.

I believe this is what you want to do. You want to predict the vertical position of the ball at any given time, t, that at time t = 0 has just struck the ground and is about to bounce back upwards with some initial velocity, yes?

If that is the case you need a reference point, let's pick the ground. The ball starts off where with respect to the ground? Just at it right? So the initial height is zero, I have no idea where you got -23.8, do not bother telling me as this is wasting my time. I am just telling you for the last time how to set up your equation. Whether you do so properly or not I care not anymore. So C should be zero. B is the initial vertical velocity the ball has just as it leaves the ground and starts heading upwards. How did you get that? You need some kind of speed sensor to find the initial velocity just after the ball strikes the ground and bounces back. That is B and it cannot be 21.9 meters/sec because the ball would travel upwards far more than about 1 meter as you say, so you need to find the correct vertical velocity the ball has when it leaves the ground and is heading upwards.

The equation works for finding the position both horizontally and vertically, but I am trying to make this as simple as possible and saying to just go with the vertical velocity. If not you need to know the angle at which the ball leaves the ground at so you can resolve the velocity components into the vertical and horizontal velocities and then use two separate equations. One to describe the position of the ball horizontally and the other vertically. The vertical equation has a constant acceleration acting on the ball - gravity. The horizontal component is unaffected by gravity and is only slowed by air resistance, and the rate of change in horizontal velocity due to air resistance (the horizontal acceleration acting on the ball in that direction) is not constant but depends on the speed of the ball itself. So let's just work with vertical velocities and positions.

So the initial velocity should be found by dropping the ball from chest height straight down and once it strikes the ground and bounces back, the moment it leaves the ground bouncing back you need to know that velocity to plug in for B.

A is just one half the acceleration due to gravity, and C is the initial position the ball is at when the experiment starts, and from your graph that very much appears to be zero if you are measuring height with respect to the ground. I do not mean to get upset with you, but you need to think through the problem yourself some to an extent, and it seems as if you are not. I have been telling you how the equation is used, and you keep replying with answers that further tell us what you are trying to do, but it is as if you are ignoring what I am telling you as to how the equation is used. I can see what you want to do.

Drop the ball from some height and let it strike the ground. It will bounce back heading upwards. Now the instant it leaves the ground heading upwards with some speed; that speed is what you need to know. It is B. Since the ball started by bouncing off the ground, make it easy on yourself and call the initial height zero. So C is zero. A is just one half the acceleration due to gravity.

That's your equation and that is how you should be setting it up if you wish it to predict the position of the ball vertically with respect to the ground at any time, t, with any accuracy whatsoever.

Do you know what I mean?!

Craig