Projectile motion question (1 Viewer)

[cutemouse]

Well if a projectile does stop at top of flight, then there must be a way to calculate for HOW LONG it stops for.

Saying that it stops for an infinitely small time without justification (ie. calculations), isn't really valid. How do you know it stops for a very small time? Can you prove it?

 

[youngminii]

Well if a projectile does stop at top of flight, then there must be a way to calculate for HOW LONG it stops for.

Saying that it stops for an infinitely small time without justification (ie. calculations), isn't really valid. How do you know it stops for a very small time? Can you prove it?

Barring the fact that it's common sense
I believe it's the same as proving what I stated before. Now it's not within the scope of our syllabus, you're just meant to assume this. You could go find the proof in google, perhaps even in your (Maths) textbook, but my advice to you is to go and relearn limits in Maths if you want a better understanding.

 

[cutemouse]

Common sense, assume? I think you're getting quite desperate to justify yourself Do you think, say special relativity is 'common sense'?

I too used to think that a projectile did 'stop' at top of flight but my Physics teacher did correct me with all this, and trust me he's a very good teacher.

I don't have to find the proof, I think that onus in on you

 

[youngminii]

Yes, we are meant to assume things.
Why don't you prove to me every formula that's been given to you?
Why don't you prove to me why light is invariant?

 

[cutemouse]

That's irrelevant.

In Physics they wont ask any questions that you need to assume that the projectile stops at the top of flight.

All I'm trying to say that is Vy=0 is just a number, and that is it. The projectile doesn't stop at the top of flight.

Earth's accel. due to gravity is about 9.8ms^-2. If you graph the motion of a projectile on a velocity/time graph then it's just a straight line with a negative gradient that cuts the x-axis (the gradient of course depends on which direction you define positive). And of course the gradient would equal about 9.8ms^-2.

If you went to a planet with accel. due to gravity which was significently less (like 0.00001ms^-2), and recorded the motion of a projectile on a graph then according to you, at Vy=0, there would be a part where the gradient = 0. This of course would contradict the fact that there is constant accel. due to gravity, wouldn't it?

And I think Einstein proved that the speed of light in a vacuum is constant -- So stop saying irrelevant stuff.

 

[youngminii]

That's irrelevant.

In Physics they wont ask any questions that you need to assume that the projectile stops at the top of flight.

All I'm trying to say that is Vy=0 is just a number, and that is it. The projectile doesn't stop at the top of flight.

Earth's accel. due to gravity is about 9.8ms^-2. If you graph the motion of a projectile on a velocity/time graph then it's just a straight line with a negative gradient that cuts the x-axis (the gradient of course depends on which direction you define positive). And of course the gradient would equal about 9.8ms^-2.

If you went to a planet with accel. due to gravity which was significently less (like 0.00001ms^-2), and recorded the motion of a projectile on a graph then according to you, at Vy=0, there would be a part where the gradient = 0. This of course would contradict the fact that there is constant accel. due to gravity, wouldn't it?

And I think Einstein proved that the speed of light in a vacuum is constant -- So stop saying irrelevant stuff.

Whatever floats your boat, buddy.

 

[cutemouse]

Undeniable Eh?

 

[shaon0]

Undeniable Eh?

Seriously. Everyone's argued a point except you. You have no counter-argument.
Everyone knows the projectile stops at one point within its flight.
The projectile stops when the derivatives=0 ie. f'(x)=0

 

[cutemouse]

Okay, so you're telling me that if you go on a planet with a VERY VERY VERY VERY LITTLE gravitational force, and you throw a ball vertically for a very short distance then the velocity/time graph will not be a straight line going through the x-axis (ie where Vy=0), but it would have a part where the gradient=0 on the axis?

Going through zero on the other hand DOES NOT MEAN THE PROJECTILE STOPS!

If you TRIED to read what I posted above, then maybe you'd see that I illustrated this point.

I'll repeat that, if you're saying that a projectile stops at the top of flight, then you need to be able to tell me exactly how long it stops for. Why don't you find some experimental evident supporting your claim that a projectile should stop at top of flight? (I'll tell you why - because there isn't any, which is because it doesn't stop!).

Of course I'm speaking of theoretical values (ie. ignoring air resistance etc etc).

Boy this is fun, it's like arguing that the aether exists

The projectile stops when the derivatives=0 ie. f'(x)=0

The derivative of what, and for how long does a projectile stop then?

 

[Trebla]

The derivative of what, and for how long does a projectile stop then?

That question has been answered several times, and you have yet to discredit them. You're asking for a finite time at which the projectile "stops" at which is an extremely flawed question in itself. It's a bit like asking "how long does a ball falling to the ground, touch the ground before it bounces off the ground?". Going by your analogy the answer would be "the ball doesn't touch the ground", when it obviously does.

If the projectile didn't stop it would not have any property of v_y = 0.

 

[cutemouse]

Trebla, as I have said MANY times. I said that the projectile in a flight GOES THROUGH Vy=0. Like any other velocity, it goes through Vy=1, Vy=2, Vy=3, etc etc etc... Not disagreeing with that. But I'm saying that the projectile DOES NOT stop at Vy=0. It is just like any those other quantities. How long does the ball stop when it goes through Vy=1, Vy=2, Vy=3, etc? Obviously zero, so why should it stop when it GOES THROUGH Vy=0?

And with the ball hitting the ground, I believe that it is less practical as there are various factors. First thing is obviously if a ball is elastic then the story will be different. Secondly it gives off sound and little bit of heat energy when it hits the ground. I don't know how those factors could affect the ball.

But if we assume that the surface of the ground can withstand the full force of the ball, no sound energy or any other energy transformations occur, and the ball is rock solid (ie. not elastic), then obviously it does not stop when it hits the ground and bounced back up. However, its vertical velocity, Vy, will be zero at one point, as it GOES THROUGH Vy=0!

You've actually convinced me even more that that a ball does not stop at top of flight

 

[Trebla]

If it goes through v_y = 0, then by definition, it does "stop" at some point.

Stop does not have to mean zero velocity at a finite time.

You've misread my post about the bouncing scenario. I did not say whether the ball would stop. I said whether it would touch the ground.

Let me ask you this: "how long does the ball touch the ground?" (assume elastic collision)

 

[shaon0]

Trebla, as I have said MANY times. I said that the projectile in a flight GOES THROUGH Vy=0. Like any other velocity, it goes through Vy=1, Vy=2, Vy=3, etc etc etc... Not disagreeing with that. But I'm saying that the projectile DOES NOT stop at Vy=0. It is just like any those other quantities. How long does the ball stop when it goes through Vy=1, Vy=2, Vy=3, etc? Obviously zero, so why should it stop when it GOES THROUGH Vy=0?

And with the ball hitting the ground, I believe that it is less practical as there are various factors. First thing is obviously if a ball is elastic then the story will be different. Secondly it gives off sound and little bit of heat energy when it hits the ground. I don't know how those factors could affect the ball.

But if we assume that the surface of the ground can withstand the full force of the ball, no sound energy or any other energy transformations occur, and the ball is rock solid (ie. not elastic), then obviously it does not stop when it hits the ground and bounced back up. However, its vertical velocity, Vy, will be zero at one point, as it GOES THROUGH Vy=0!

You've actually convinced me even more that that a ball does not stop at top of flight

I think your getting confused with Hyperreal Number system and the Normal Number System because the the Normal Number system only gives a finite value for distance.
For an infinitely small amount of time, Vy = 0.

Think about a rollercoaster, At the highest point, you stop.

 

[cutemouse]

If it goes through v_y = 0, then by definition, it does "stop" at some point.

Stop does not have to mean zero velocity at a finite time.

Then your definition of 'stop' is different to mine. When I say stop, I mean that the projectile has no velocity for a certain period of time.

Let me ask you this: "how long does the ball touch the ground?" (assume elastic collision)

What you're trying to point out is something a little bit different. First of all, the projectile has a velocity (that isn't zero) before it hits the ground. Second of all, it undergoes acceleration, whilst the projectile is accelerating, so that obviously makes it a little different. So it is not the same as the ball at top of the flight, and your question cannot be proven easily, as you cannot get an elastic collision.

In practise, the projectile/ball may stop at bottom of flight as you cannot get an elastic collision, as I said before. But in theory, the projectile's velocity shouldn't be equal to zero at any point as it would be an instantaneous of direction of velocity (ie. from -v to v or vice versa)

So, the answer to your question: I don't know. But it's a little different from what I'm trying to say.

 

[shaon0]

Then your definition of 'stop' is different to mine. When I say stop, I mean that the projectile has no velocity for a certain period of time.


What you're trying to point out is something a little bit different. First of all, the projectile has a velocity (that isn't zero) before it hits the ground. Second of all, it undergoes acceleration, whilst the projectile is accelerating, so that obviously makes it a little different. So it is not the same as the ball at top of the flight, and your question cannot be proven easily, as you cannot get an elastic collision.

In practise, the projectile/ball may stop at bottom of flight as you cannot get an elastic collision, as I said before. But in theory, the projectile's velocity shouldn't be equal to zero at any point as it would be an instantaneous of direction of velocity (ie. from -v to v or vice versa)

So, the answer to your question: I don't know. But it's a little different from what I'm trying to say.

Wow, now you're just becoming incoherent lol

 

[Trebla]

What you're trying to point out is something a little bit different. First of all, the projectile has a velocity (that isn't zero) before it hits the ground. Second of all, it undergoes acceleration, whilst the projectile is accelerating, so that obviously makes it a little different. So it is not the same as the ball at top of the flight, and your question cannot be proven easily, as you cannot get an elastic collision.

A projectile also undergoes acceleration...we are not concerned with the acceleration after it collides.

In practise, the projectile/ball may stop at bottom of flight as you cannot get an elastic collision, as I said before. But in theory, the projectile's velocity shouldn't be equal to zero at any point as it would be an instantaneous of direction of velocity (ie. from -v to v or vice versa)

So, the answer to your question: I don't know. But it's a little different from what I'm trying to say.

Well, using your idea that you cannot get elastic collision, then you cannot get a perfect projectile because of air resistance. A projectile is a theoretical model to explain a certain type of motion as is an elastic collision.

You said that the projectile's velocity shouldn't equal zero at any point. Well how can it go from negative velocity to positive velocity without "passing zero"? Change in velocity is continuous, so it can't exactly jump from negative to positive without passing zero.

The intention of the question was to see if you understood the concept of infinitesimal time. The ball DOES touch the ground, but for how long? An infinitely small amount of time. It does not merely "pass through" the ground. You acknowledge that it does touch the ground, so that means at some point, the velocity hits zero as it changes direction. This is analogous to your question of "how long does the projectile stop?" The answer is also an infinitely small amount of time.

If your definition of stop refuses to include infinitesimal times (which finite times approach anyway when they are sufficiently small) then your original question was flawed to begin with.

 

[cutemouse]

Well, using your idea that you cannot get elastic collision, then you cannot get a perfect projectile because of air resistance. A projectile is a theoretical model to explain a certain type of motion as is an elastic collision.

If there is no air (like in vacuum) then there is no air resistance.

You said that the projectile's velocity shouldn't equal zero at any point. Well how can it go from negative velocity to positive velocity without "passing zero"? Change in velocity is continuous, so it can't exactly jump from negative to positive without passing zero.

That is going too far into Physics/Maths, which is not what I intended. But it is different to a ball accelerating due to gravity and then falling back.

The intention of the question was to see if you understood the concept of infinitesimal time. The ball DOES touch the ground, but for how long? An infinitely small amount of time. It does not merely "pass through" the ground. You acknowledge that it does touch the ground, so that means at some point, the velocity hits zero as it changes direction. This is analogous to your question of "how long does the projectile stop?" The answer is also an infinitely small amount of time.

My point was that, the time that it touches the ground is undefined. Therefore you cannot make the conclusion/assume that a projectile stops. If the time it stops for is undefined, then it doesn't stop as you cannot give a quantitative measurement of how long it stops for.

If your definition of stop does not include infinitesimal times (which finite times approach anyway when they are sufficiently small) then your original question was flawed to begin with.

My question? I was responding to:

The vertical component is 0 for a moment

 

[Trebla]

My question? I was responding to:

I meant your question "How long does a projectile stop for?"

That is going too far into Physics/Maths, which is not what I intended. But it is different to a ball accelerating due to gravity and then falling back.

What on Earth are you talking about? How are they different? The velocity passes through zero when it touches the ground as it does when it is at maximum height. The only way to get it to you is using mathematical arguments under the models, because you refuse to accept anything else.

My point was that, the time that it touches the ground is undefined. Therefore you cannot make the conclusion/assume that a projectile stops. If the time it stops for is undefined, then it doesn't stop as you cannot give a quantitative measurement of how long it stops for.

Just because it is not well defined (but even so, it is in fact extremely close to being well defined), doesn't mean it does not exist. If you recall Year 11 differential calculus, the gradient dy/dx is not "well defined" as you put it because it is a LIMIT of a line joining two points on a curve and these two points merging into one point to give the gradient of a curve at a point. It is a perfect example of use of infinitesimals. Does that mean the gradient of a curve doesn't exist? Are turning points non-existent?

You say that the time a projectile stops is undefined so it therefore does not stop, but you also say that "the time is touches the ground is undefined", so by the same analogy the ball does not touch the ground. You know and acknowledged that it obviously touches the ground, so you've contradicted yourself.

 

[cutemouse]

No, I haven't contradicted myself...

According to you the projectile stops for an infinitesimal amount of time, which is undefined at the top of flight (ie. at Vy=0).

At various points towards top of flight the projectile has a velocity of say, Vy=2 and then this changes to Vy=1 which changes to Vy=0, Vy=-1, and Vy=-2, and so on. The rate of change of velocity is constant.

Then, keeping that in mind, you're saying that the projectile remains at Vy=0 for an infinitesimal of time. But then shouldn't the projectile then remain at say, Vy=2 for an infinitesimal amount of time aswell? But everyone would say that the projectile wouldn't stop at Vy=2. My point is that 'infinitesimal' is not a value (its undefined) and that we do not consider a projectile to stop in any parts of its flight.

 

[cutemouse]

What on Earth are you talking about? How are they different?

In your case, it's collision, NOT constant acceleration (or rate of change of velocity), which makes it different.

The velocity passes through zero when it touches the ground as it does when it is at maximum height. The only way to get it to you is using mathematical arguments under the models, because you refuse to accept anything else.

It goes through zero yes, but if the velocity was say, v=36m/s downwards initially, then according to you it should go through v=35, 34, 33, ... ,-36. It doesn't mean that it stops at any of those velocities through!