Parabolic path of ball-speed before hitting the ground

[negation]

Homework Statement

You toss a ball straight up at 7.3ms^-1; it leaves your hand at 1.5m above the floor.
a) Find its velocity before it hits the floor
b) Suppose ball is tossed straight down at 7.3ms^-1, what is the final velocity just before it hits the floor.
c) When would the second ball hit the floor.

Homework Equations

None

The Attempt at a Solution

 

[Curious3141]

Homework Statement

You toss a ball straight up at 7.3ms^-1; it leaves your hand at 1.5m above the floor.
a) Find its velocity before it hits the floor
b) Suppose ball is tossed straight down at 7.3ms^-1, what is the final velocity just before it hits the floor.
c) When would the second ball hit the floor.

Homework Equations

None

The Attempt at a Solution

View attachment 65020

Don't have time for a long post tonight, but it just seems to me that you're doing a lot of unnecessary work here - and it's also easy to make mistakes when overworking a problem.

You've posted a lot of kinematic problems, and I've helped you with some of them. I wanted to say something before, but I'll say it now: in kinematics, it's always good to start by writing down the principal variables, and then listing the ones you're given, then finally what you need to determine. I'm not sure what form of equations you've learnt, but this is what I know:

$v = u + at$
$v^2 = u^2 + 2as$
$s = ut + \frac{1}{2}at^2$

Those are the three you *have* to remember.

Then there are these two which are easily derived from the others, but they're good to remember as well:

$s = vt - \frac{1}{2}at^2$
$s = \frac{1}{2}(u+v)(t)$

The variables are a,u,v,t and s. Start by listing those out and writing what you know.

For instance, in this question's part a), you're given u (=+7.3),a (=-g) and s (=-1.5). You're asked to determine v. Which equation would you select for a one step solution? Proceed like that.

Would you expect the answers to parts a) and b) to be different? Even without working them out?

 

[negation]

Don't have time for a long post tonight, but it just seems to me that you're doing a lot of unnecessary work here - and it's also easy to make mistakes when overworking a problem.

You've posted a lot of kinematic problems, and I've helped you with some of them. I wanted to say something before, but I'll say it now: in kinematics, it's always good to start by writing down the principal variables, and then listing the ones you're given, then finally what you need to determine. I'm not sure what form of equations you've learnt, but this is what I know:

$v = u + at$
$v^2 = u^2 + 2as$
$s = ut + \frac{1}{2}at^2$

Those are the three you *have* to remember.

Then there are these two which are easily derived from the others, but they're good to remember as well:

$s = vt - \frac{1}{2}at^2$
$s = \frac{1}{2}(u+v)(t)$

The variables are a,u,v,t and s. Start by listing those out and writing what you know.

For instance, in this question's part a), you're given u (=+7.3),a (=-g) and s (=-1.5). You're asked to determine v. Which equation would you select for a one step solution? Proceed like that.

Would you expect the answers to parts a) and b) to be different? Even without working them out?

Vf^2 - vi^2 =delta y

I know man, been trying to get a head start or next semester's unit during this summer break. Just finished foundation year in uni.
There's a total of 86 questions. I've finished 83 question. Will move on to the next topic when done.

EDIT: Same because vi are both the same weather the ball is thrown straight up from ground of if the ball is tossed at the same speed as if it were thrown straight up from ground.

 

[Curious3141]

Vf^2 - vi^2 =delta y
Different, because vf is dependent on vi. Delta y is a constant and so is acceleration due to gravity.
I know man, been trying to get a head start or next semester's unit during this summer break. Just finished foundation year in uni.
There's a total of 86 questions. I've finished 83 question. Will move on to the next topic when done.

No worries, you're doing quite well. But my gentle suggestion is to try and be a little more systematic, as I mentioned in my previous post. This question doesn't actually need you to solve a quadratic at all, if you work through it in the order given.

I guess you meant $v_f^2 - v_i^2 = 2a\Delta{y}$, which is the form you learned it in? That's fine, and if you apply it properly there shouldn't be an issue.

Remember what I said before about sign conventions? Easiest to let "up" be positive and "down" be negative, because that's intuitive to most people.

So in part a), what's $v_i$?

And in part b), what's $v_i$?

What are $a$ and $\Delta{y}$ in each case? Are they the same?

So what is $v_f$ in parts a) and b)? The same or different?

Remember that when you're given a constant acceleration, the kinematic equations can always be applied directly. Even if the trajectory involves going up (because of an initial upward throw), then down (because of gravity acting downward), you don't actually have to calculate the maximum height, time taken to reach that, etc. unless you're actually asked those things. Immediately after leaving the hand, the only force that acts on the object is gravity, and that's constant, so the kinematic equations apply to the whole trajectory. No need to break it up into an 'up' and 'down' stage. Just be careful with your signs and interpret any times you get intelligently, and you should be fine.

 

[negation]

No worries, you're doing quite well. But my gentle suggestion is to try and be a little more systematic, as I mentioned in my previous post. This question doesn't actually need you to solve a quadratic at all, if you work through it in the order given.

I guess you meant $v_f^2 - v_i^2 = 2a\Delta{y}$, which is the form you learned it in? That's fine, and if you apply it properly there shouldn't be an issue.

Remember what I said before about sign conventions? Easiest to let "up" be positive and "down" be negative, because that's intuitive to most people.

So in part a), what's $v_i$?

And in part b), what's $v_i$?

What are $a$ and $\Delta{y}$ in each case? Are they the same?

So what is $v_f$ in parts a) and b)? The same or different?

Remember that when you're given a constant acceleration, the kinematic equations can always be applied directly. Even if the trajectory involves going up (because of an initial upward throw), then down (because of gravity acting downward), you don't actually have to calculate the maximum height, time taken to reach that, etc. unless you're actually asked those things. Immediately after leaving the hand, the only force that acts on the object is gravity, and that's constant, so the kinematic equations apply to the whole trajectory. No need to break it up into an 'up' and 'down' stage. Just be careful with your signs and interpret any times you get intelligently, and you should be fine.

I just edited a few seconds before your new post.
vi in both circumstances are both the same whether ball is thrown up at 7ms^-1 or thrown down at 7ms^-1.

My mathematical workings reveals the above is true: v_f = 9.09ms^-1
Yes and I did allowed g to be negative.
vf^2 - vi^2 = 2(-g)(delta y)
I think I see where the confusion came from. I've always been using negative g but in the event where a numerical g is not required, I would write-suppose any kinematics equation with a g variable as vf^2 - vi^2 = 2(g)(delta y) where the g would, if a numerical value is required, the appropriate sign would be input. So where a numerical value of g is not required, it would appear as thought I took g to be positive.

 

[negation]

Also, do you mind explaining the concept and derivation behind sqrt 2d/g?

 

[Curious3141]

EDIT: Same because vi are both the same weather the ball is thrown straight up from ground of if the ball is tossed at the same speed as if it were thrown straight up from ground.

You edited after my post. Anyway, this correct. $v_i$ only differs in sign, the magnitude is the same. When you square a positive or negative number of the same magnitude, the result is the same.

Another way to look at it (just for interest) is conservation of energy. When you throw the ball up, you impart a certain kinetic energy to it. As it goes upward, this energy gets progressively converted into gravitational potential energy that exists in the field between the ball and the Earth. At the max height, the ball is momentarily at rest, and all the energy is in the form of gravitational potential energy (zero kinetic energy). As it comes down again the reverse process occurs, and there is conversion of potential energy to kinetic energy. As the ball passes its former launching height traveling downward, it will have the exact same speed going down as it had going up (even though the direction is reversed). This is a consequence of the conservation of energy.

A question to test your understanding: what "real world" condition can affect the trajectory so that this doesn't apply (and the ball will have a different speed going down than it did being thrown up)?

Getting back to the kinematics treatment, when you calculate $v_f$ you need to take the square root of something. Remember there is a positive and a negative square root, and which you take depends on the physical interpretation. Which is correct here?

 

[Curious3141]

Also, do you mind explaining the concept and derivation behind sqrt 2d/g?

Do you mean the kinematic equation $v_f^2 -v_i^2 = 2as$? Better to clarify first.

 

[negation]

You edited after my post. Anyway, this correct. $v_i$ only differs in sign, the magnitude is the same. When you square a positive or negative number of the same magnitude, the result is the same.

Another way to look at it (just for interest) is conservation of energy. When you throw the ball up, you impart a certain kinetic energy to it. As it goes upward, this energy gets progressively converted into gravitational potential energy that exists in the field between the ball and the Earth. At the max height, the ball is momentarily at rest, and all the energy is in the form of gravitational potential energy (zero kinetic energy). As it comes down again the reverse process occurs, and there is conversion of potential energy to kinetic energy. As the ball passes its former launching height traveling downward, it will have the exact same speed going down as it had going up (even though the direction is reversed). This is a consequence of the conservation of energy.

A question to test your understanding: what "real world" condition can affect the trajectory so that this doesn't apply (and the ball will have a different speed going down than it did being thrown up)?

Getting back to the kinematics treatment, when you calculate $v_f$ you need to take the square root of something. Remember there is a positive and a negative square root, and which you take depends on the physical interpretation. Which is correct here?

±9.09ms^-1 but in ascribing downwards to imply a negative direction and upwards, positive, v_i = -9.09ms^-1

Are you familiar with solving this problem using the concept of limit?

 

[Curious3141]

v_f = 9.09ms^-1

Should $v_f$ be positive? See my previous post.

Yes and I did allowed g to be negative.
vf^2 - vi^2 = 2(-g)(delta y)
I think I see where the confusion came from. I've always been using negative g but in the event where a numerical g is not required, I would write-suppose any kinematics equation with a g variable as vf^2 - vi^2 = 2(g)(delta y) where the g would, if a numerical value is required, the appropriate sign would be input. So where a numerical value of g is not required, it would appear as thought I took g to be positive.

It's OK, whether you want to assign $g = -9.81ms^{-2}$ all the time, or you want to have the convention where $g$ just represents the numerical value, and you put the sign accordingly. Whatever it is, consistency is the key.

 

[negation]

Another way to look at it (just for interest) is conservation of energy. When you throw the ball up, you impart a certain kinetic energy to it. As it goes upward, this energy gets progressively converted into gravitational potential energy that exists in the field between the ball and the Earth. At the max height, the ball is momentarily at rest, and all the energy is in the form of gravitational potential energy (zero kinetic energy). As it comes down again the reverse process occurs, and there is conversion of potential energy to kinetic energy. As the ball passes its former launching height traveling downward, it will have the exact same speed going down as it had going up (even though the direction is reversed). This is a consequence of the conservation of energy.

Yes, I'm aware of this concept. Graphically, it takes on a sin graph, yes?

 

[negation]

Should $v_f$ be positive? See my previous post.


It's OK, whether you want to assign $g = -9.81ms^{-2}$ all the time, or you want to have the convention where $g$ just represents the numerical value, and you put the sign accordingly. Whatever it is, consistency is the key.

It should be negative since the v_f prior to hitting the ground is traveling in a downwards direction.

 

[Curious3141]

Yes, I'm aware of this concept. Graphically, it takes on a sin graph, yes?

What, exactly, follows a sine graph? Not everything that is periodic is a sine relationship. Some physical systems are periodic in a sinusoidal sense, like an oscillating idealised, friction-free, drag-free spring-mass system or (approximately) a pendulum exhibiting planar oscillations of small angular amplitude. But this system is not periodic, it just "retraces its path" the one time. And I can't see any relationship to the sine function.

Not trying to knock you down, but one needs to be rigorous in one's statements.

And what do you mean by solving with a limit concept? Not familiar with this for kinematics problems.

 

[negation]

What, exactly, follows a sine graph? Not everything that is periodic is a sine relationship. Some physical systems are periodic in a sinusoidal sense, like an oscillating idealised, friction-free, drag-free spring-mass system or (approximately) a pendulum exhibiting planar oscillations of small angular amplitude. But this system is not periodic, it just "retraces its path" the one time. And I can't see any relationship to the sine function.

Not trying to knock you down, but one needs to be rigorous in one's statements.

Even if assuming acceleration due to gravity is 0ms^-1 and there exists no resistance of any sort?
Wouldn't the object then reach a height, Δy, then back to where the ground, and repeats an infinite loop?

And what do you mean by solving with a limit concept? Not familiar with this for kinematics problems.

What happens as the limit,t, tends towards the quadratic root t = 1.6...

 

[Curious3141]

Even if assuming acceleration due to gravity is 0ms^-1 and there exists no resistance of any sort?
Wouldn't the object then reach a height, Δy, then back to where the ground, and repeats an infinite loop?

If acceleration was 0 (no gravitational influence), and there was no drag or any other force acting on it, the object would either remain at rest (if not thrown) or travel at a constant velocity after the initial throw. Same speed, same direction, forever. No parabolic path. It would never even reach the ground.

And whether there is gravity or not, why would the object ever go in an infinite loop? With gravity, the object would hit the ground and stay there. Unless it had elastic properties.

What happens as the limit,t, tends towards the quadratic root t = 1.6...

Not clear at all, sorry.

 

[negation]

If acceleration was 0 (no gravitational influence), and there was no drag or any other force acting on it, the object would either remain at rest (if not thrown) or travel at a constant velocity after the initial throw. Same speed, same direction, forever. No parabolic path. It would never even reach the ground.

And whether there is gravity or not, why would the object ever go in an infinite loop? With gravity, the object would hit the ground and stay there. Unless it had elastic properties.



Not clear at all, sorry.

Thanks. Informative. I missed one of your question earlier. Frictional force would result in the ball returning to a lower height than the max height achieved when it was thrown at the outset.

 

[Curious3141]

Thanks. Informative. I missed one of your question earlier. Frictional force would result in the ball returning to a lower height than the max height achieved when it was thrown at the outset.

Not really. Friction with the air (generally called drag) would result in the ball having a lower speed on its downward trajectory compared to its upward trajectory at the same height (horizontal level). Not sure why you think it will return to a lower height - the ball always hits a maximum height, drops to the ground and stays put there. Unless it bounces (the elastic property I mentioned earlier) - only then does it make sense to talk about returning to another height.

 

[negation]

Not really. Friction with the air (generally called drag) would result in the ball having a lower speed on its downward trajectory compared to its upward trajectory at the same height (horizontal level). Not sure why you think it will return to a lower height - the ball always hits a maximum height, drops to the ground and stays put there. Unless it bounces (the elastic property I mentioned earlier) - only then does it make sense to talk about returning to another height.

I would really appreciate if you could shed some light on this:

In part (a): vi = 7.3ms^-1 while delta y = -1.5m
Why is delta y= -1.5m? Is there a rational for not measuring the displacement from ground to max height as opposed to max height to ground?

 

[negation]

part (c) When would the second ball hit the floor:

Why am I getting a negative t?

The displacement has to be negative because yf < yi and indeed true when yf = 0m and yi = -1.5m above ground as the ball is falling towards the ground and we denote downwards as negative

 

[Curious3141]

I would really appreciate if you could shed some light on this:

In part (a): vi = 7.3ms^-1 while delta y = -1.5m
Why is delta y= -1.5m? Is there a rational for not measuring the displacement from ground to max height as opposed to max height to ground?

It's not max height (at least in the case of the ball being thrown upward), it's just the height of the hand above the floor.

No particular rationale, you get to choose your convention, as long as you're consistent. If you want $\Delta{y}$ to be +1.5m, then your convention is that "up" is negative and "down" is positive. In which case, your acceleration will be +9.81 m/s^2 and your initial upward throw (for part a) will be with a velocity of -7.3 m/s.

 

[Curious3141]

part (c) When would the second ball hit the floor:

View attachment 65041

Why am I getting a negative t?

The displacement has to be negative because yf < yi and indeed true when yf = 0m and yi = -1.5m above ground as the ball is falling towards the ground and we denote downwards as negative

Recheck your convention. Initial y (at hand level) = 0. Final y (at ground) = -1.5m. Alternatively, you could assign the ground to be zero, and the hand to be +1.5m, in which case, the displacement (final minus initial) is still -1.5m. What you wrote is self-contradictory, because you stated yf < yi, but 0 is greater than -1.5m.

You can do this much more simply without the quadratic using the result from part b).

 

[negation]

It's not max height (at least in the case of the ball being thrown upward), it's just the height of the hand above the floor.

No particular rationale, you get to choose your convention, as long as you're consistent. If you want $\Delta{y}$ to be +1.5m, then your convention is that "up" is negative and "down" is positive. In which case, your acceleration will be +9.81 m/s^2 and your initial upward throw (for part a) will be with a velocity of -7.3 m/s.

What I meant was in part (a), we took vi = 7.3ms^-1 because we denote upwards to be positive. In this frame, wouldn't yi = 1.5m and yf = max height achieved by ball? Why is yf - yi then negative?
In short, why are we measuring only the distance traveled by the ball as it falls back to the ground?

 

[negation]

Recheck your convention. Initial y (at hand level) = 0. Final y (at ground) = -1.5m. Alternatively, you could assign the ground to be zero, and the hand to be +1.5m, in which case, the displacement (final minus initial) is still -1.5m. What you wrote is self-contradictory, because you stated yf < yi, but 0 is greater than -1.5m.

You can do this much more simply without the quadratic using the result from part b).

Yes you're right. It became much clearer if I visualized a y-coordinate as a reference.
I know: vf = vi+at? I prefer the quadratic though. It gives me a global picture as to what is occurring.

 

[Curious3141]

What I meant was in part (a), we took vi = 7.3ms^-1 because we denote upwards to be positive. In this frame, wouldn't yi = 1.5m and yf = max height achieved by ball? Why is yf - yi then negative?
In short, why are we measuring only the distance traveled by the ball as it falls back to the ground?

Why should the max height even enter into consideration? You're not asked to work it out.

As I said before, there is *no need* to bother about this. All you need in a kinematics problem is to consider the initial displacement (hand, in this case) and the final displacement (ground, in this case). The ball can do *anything* in the middle (subject to the constraints of a constant acceleration, otherwise the equations don't apply). But you don't have to worry about it.

If you're asked to calculate the max height, THEN you have to worry about it, and you do that by setting the final velocity to zero (because the ball momentarily comes to rest at max height). Then you use the equation relating initial vel, final vel, acceleration and displacement.

EDIT: I see where you might be confused. You might want to review your notes very carefully about the difference between a "distance" and a "displacement". They're related, but quite different, really. Displacement is all that matters in the kinematic equations. But if you want to calculate total distance traveled by the ball as it goes up then goes down to hit the ground, you definitely need to divide the trajectory into two parts and calculate each displacement separately, then add the absolute values.

 

[negation]

Why should the max height even enter into consideration? You're not asked to work it out.

As I said before, there is *no need* to bother about this. All you need in a kinematics problem is to consider the initial displacement (hand, in this case) and the final displacement (ground, in this case). The ball can do *anything* in the middle (subject to the constraints of a constant acceleration, otherwise the equations don't apply). But you don't have to worry about it.

If you're asked to calculate the max height, THEN you have to worry about it, and you do that by setting the final velocity to zero (because the ball momentarily comes to rest at max height). Then you use the equation relating initial vel, final vel, acceleration and displacement.

EDIT: I see where you might be confused. You might want to review your notes very carefully about the difference between a "distance" and a "displacement". They're related, but quite different, really. Displacement is all that matters in the kinematic equations. But if you want to calculate total distance traveled by the ball as it goes up then goes down to hit the ground, you definitely need to divide the trajectory into two parts and calculate each displacement separately, then add the absolute values.

I worded it quite poorly in haste.
Given the equation vf^2 - vi^2 = 2(-g)(yf-yi):
Using the equation above, am I right to imply that the inputs must be variables occurring within a "specific frame" such as either all variables involved as the ball goes upwards or downwards but never an admixture of variables from 2 occurring frames.
As the ball is tossed, vi = +, g = - and yf >yi but it seems that delta y = - and this is true iff
yf < yi.
From this, and using backwards induction, I deduced that we are interested in the displacement as the ball falls towards the ground.
Therefore, I'm saying isn't it strange that if vi is + upwards then in this occurring frame yf must be a particular positive max height (not that we are interested in max height but I'm expounding on a point)? But it appears as though we split and "mixed" the event of the ball going up and down.
I was under the assumption that if vi = + up, then any unknown in that equation would only tells us everything about what is happening as the ball goes up.

Edit: I did not saw your edit as I was already replying in response to your quote.

 

[Curious3141]

I worded it quite poorly in haste.
Given the equation vf^2 - vi^2 = 2(-g)(yf-yi):
Using the equation above, am I right to imply that the inputs must be variables occurring within a "specific frame" such as either all variables involved as the ball goes upwards or downwards but never an admixture of variables from 2 occurring frames.
As the ball is tossed, vi = +, g = - and yf >yi but it seems that delta y = - and this is true iff
yf < yi.
From this, and using backwards induction, I deduced that we are interested in the displacement as the ball falls towards the ground.
Therefore, I'm saying isn't it strange that if vi is + upwards then in this occurring frame yf must be a particular positive max height (not that we are interested in max height but I'm expounding on a point)? But it appears as though we split and "mixed" the event of the ball going up and down.
I was under the assumption that if vi = + up, then any unknown in that equation would only tells us everything about what is happening as the ball goes up.

Edit: I did not saw your edit as I was already replying in response to your quote.

What you wrote is confusing. Basically, the take home message is this: it's all one "frame" (or convention) as long as there is a constant acceleration throughout. Just choose a convenient convention and stick to it! It doesn't matter which way the body is moving or whether it changes direction halfway. All the equations apply throughout.

Sometimes, after solving, you may get more than one answer (usually a maximum of two, since we're dealing with quadratics, at most). Often one answer is physically impossible (like a negative time). But sometimes, both answers are physically plausible. That's when you have to decide, based on your understanding of the physical situation, which answer is the correct one.

 

[negation]

What you wrote is confusing. Basically, the take home message is this: it's all one "frame" (or convention) as long as there is a constant acceleration throughout. Just choose a convenient convention and stick to it! It doesn't matter which way the body is moving or whether it changes direction halfway. All the equations apply throughout.

I'll contemplate about it. Once I understand fully the convention, I should be able to move on to vectors.