Projectile motion question (2 Viewers)

[Trebla]

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.

I never said the projectile stops at say v_y = 2. You just made that up. You even acknowledged that according to me it stops at v_y = 0, so why on Earth did you suddenly change your mind and say that I supposedly said it stops at all points on the projectile?

It stops at v_y = 0 and the duration it takes the value v_y = 0 is an infinitesimal time.

You're confusing yourself. I never said it stops at any other part of its flight other than when it hits v_y = 0. You seem to think that infinitesimal time = stop, which is not what I meant at all.

It does take v_y = 2 at an infinitesimal time as well, but I meant that specifically at v_y = 0 is the point where it "stops" and its duration is an infinitesimal time which was in answer to your question "how long does the projectile stop for?".

Again you raised the idea that "infinitesimal is not a value and is undefined", well you've just raised the same thing (that things with infinitesimals are non-existent) without even looking or disproving my argument against it:

"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?"

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

Again, you're missing the point. We're only concerned with the idea of infinitesimal time, which is applicable to both cases (stopping and touching the ground). We are not analysing forces or anything like that, we're looking at the passing of time.

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!

It stops at v = 0!!!!

Also, you haven't really defended why you think you did not contradict yourself in:

"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]

I never said the projectile stops at say v_y = 2. You just made that up. You even acknowledged that according to me it stops at v_y = 0, so why on Earth did you suddenly change your mind and say that I supposedly said it stops at all points on the projectile?

It stops at v_y = 0 and the duration it takes the value v_y = 0 is an infinitesimal time.

You're confusing yourself. I never said it stops at any other part of its flight other than when it hits v_y = 0. You seem to think that infinitesimal time = stop, which is not what I meant at all.

Read my post again, I said that the rate of change of velocity if CONSTANT. Therefore, the rate that the velocity changes from Vy=2 to Vy=1 is the same as the rate that the velocity changes from Vy=1 to Vy=0 and from Vy=0 to Vy=-1 and so on. You're saying that the projectile stops at Vy=0. If it does, then shouldn't it remain at all the other velocities for an "infinitesimal" amount of time aswell? (since rate of change of velocities are the same) I was trying to point out that your thinking was flawed.

Again you raised the idea that "infinitesimal is not a value and is undefined", well you've just raised the same thing (that things with infinitesimals are non-existent) without even looking or disproving my argument against it:

"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?"

I don't recall that from Year 11 I'm afraid and I have no idea about the stuff that 2 points merge on one line (???)

Again, you're missing the point. We're only concerned with the idea of infinitesimal time, which is applicable to both cases (stopping and touching the ground). We are not analysing forces or anything like that, we're looking at the passing of time.

Yes, but that case of a COLLISION is not constant acceleration (ie. the rate of change of velocity is not constant when it changes direction)

It stops at v = 0!!!!

How come? According to you, since it goes from say, v=36 to v=-36, it goes through Vy=0 so it stops for an infinitesimal amount of time. But then shouldn't it also go through v=35, v=34,..., v=-35, v=-36, and thus stop at those velocities for an infinitesimal amount of time aswell? Clearly your 'theory' is flawed

Also, you haven't really defended why you think you did not contradict yourself in:

"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."

You're comparing apples with oranges. The ball hitting the ground is not constant acceleration. It does not relate to a projectile at the top of flight. I never said that I knew anything about what happens when a ball hits the ground. You're the one that brought that up.

 

[Trebla]

Read my post again, I said that the rate of change of velocity if CONSTANT. Therefore, the rate that the velocity changes from Vy=2 to Vy=1 is the same as the rate that the velocity changes from Vy=1 to Vy=0 and from Vy=0 to Vy=-1 and so on. You're saying that the projectile stops at Vy=0. If it does, then shouldn't it remain at all the other velocities for an "infinitesimal" amount of time aswell? (since rate of change of velocities are the same)

Where is the flaw to that argument? What you are basically saying is that a fundamental theorem of calculus is flawed. For an infinitesimal amount of time, it remains at v_y = 0. When it hits v_y = 1, that means that for an infinitesimal amount of time it travels at 1 m/s (not 0), and when it hits v_y = 2 it travels at 2 m/s at that moment (not 0) etc and for an infinitesimal amount of time it hits v_y = 0 which by definition stops momentarily as it is 0 m/s at that moment. Again I'll reiterate: infinitesimal time =/= stop.

I don't recall that from Year 11 I'm afraid and I have no idea about the stuff that 2 points merge on one line (???)

Then I suggest you look back on how to derive the formula from first principles to calculate the derivative of a function. This is probably the root of your misunderstanding.

Yes, but that case of a COLLISION is not constant acceleration (ie. the rate of change of velocity is not constant when it changes direction)

What does constant or non-constant acceleration have to do with infinitesimal time? You can't change time passing, so the idea of infinitesimal time is equally applicable in both scenarios regardless of the nature of acceleration. Again, you are diverting beside the point.

How come? According to you, since it goes from say, v=36 to v=-36, it goes through Vy=0 so it stops for an infinitesimal amount of time. But then shouldn't it also go through v=35, v=34,..., v=-35, v=-36, and thus stop at those velocities for an infinitesimal amount of time aswell? Clearly your 'theory' is flawed

STOP means v = 0. You are confusing the notion of stop with how long it actually stops. Clearly you misunderstand the theory (not to mention the fundamentals of calculus) and thus your assertions are extremely flawed. At v = 35 the speed is 35 m/s (not zero) and it travels at 35 m/s for an infinitesimal amount of time. This does not mean it stops, it means it travels at 35 m/s at that moment in time.

It stops only when you have the special case of v = 0 which means at that moment in time is goes to 0 m/s i.e. it stops for a moment! Again, you've misunderstood me. I'll reiterate again: infinitesimal time =/= stop.

The DURATION it takes v = 0 (note that we are talking about TIME to describe the stop, NOT the notion of stopping itself) is an infinitely small amount of time.

You're comparing apples with oranges. The ball hitting the ground is not constant acceleration. It does not relate to a projectile at the top of flight. I never said that I knew anything about what happens when a ball hits the ground. You're the one that brought that up.

They are related by the concept of infinitely small time intervals (which is something you don't seem to understand). Acceleration has nothing to do with it.

 

[cutemouse]

Where is the flaw to that argument? What you are basically saying is that a fundamental theorem of calculus is flawed. For an infinitesimal amount of time, it remains at v_y = 0. When it hits v_y = 1, that means that for an infinitesimal amount of time it travels at 1 m/s (not 0), and when it hits v_y = 2 it travels at 2 m/s at that moment (not 0) etc and for an infinitesimal amount of time it hits v_y = 0 which by definition stops momentarily as it is 0 m/s at that moment. Again I'll reiterate: infinitesimal time =/= stop.

Here we go again... -_-

Then I suggest you look back on how to derive the formula from first principles to calculate the derivative of a function. This is probably the root of your misunderstanding.

I know how to do this, thank you very much.

Again, you are diverting beside the point.

No I'm not.

STOP means v = 0. You are confusing the notion of stop with how long it actually stops. Clearly you misunderstand the theory (not to mention the fundamentals of calculus) and thus your assertions are extremely flawed. At v = 35 the speed is 35 m/s (not zero) and it travels at 35 m/s for an infinitesimal amount of time. This does not mean it stops, it means it travels at 35 m/s at that moment in time.

So when a ball hits the ground at a velocity, v=36m/s, its velocity changes to v=35, v=34, .. then v=-35 for an infinitesimal amount of time and then its velocity reaches v=-36m/s? If its velocity reaches zero (ie stops) and stays there, then where does the 'delayed' kinetic energy come from to push it back upto v=-36m/s?

It's evident that I'm not going to change my mind about this and that we're going in circles. You believe what you want, and I'll believe what I want. But the facts will always remain same.

jm01 out. *unsubscribes thread*

 

[Trebla]

So when a ball hits the ground at a velocity, v=36m/s, its velocity changes to v=35, v=34, .. then v=-35 for an infinitesimal amount of time and then its velocity reaches v=-36m/s? If its velocity reaches zero (ie stops) and stays there, then where does the 'delayed' kinetic energy come from to push it back upto v=-36m/s?

It's evident that I'm not going to change my mind about this and that we're going in circles. You believe what you want, and I'll believe what I want. But the facts will always remain same.

jm01 out. *unsubscribes thread*

At v = 0, kinetic energy is zero because at that point all of it has been converted into potential energy. As soon as v becomes non-zero, the potential energy is converted back into kinetic energy in which the ball goes back up. The time it takes for kinetic energy to be zero is an infinitely small amount of time...

 

[tacogym27101990]

i think jm01 has been thoroughly raped by trebla

 

[shaon0]

Here we go again... -_-


I know how to do this, thank you very much.


No I'm not.


So when a ball hits the ground at a velocity, v=36m/s, its velocity changes to v=35, v=34, .. then v=-35 for an infinitesimal amount of time and then its velocity reaches v=-36m/s? If its velocity reaches zero (ie stops) and stays there, then where does the 'delayed' kinetic energy come from to push it back upto v=-36m/s?

It's evident that I'm not going to change my mind about this and that we're going in circles. You believe what you want, and I'll believe what I want. But the facts will always remain same.

jm01 out. *unsubscribes thread*

Its ok to admit you are wrong. Noone thinks less of you but i did get what you were kinda trying to say, it was basically what Trebla pointed out.

 

[youngminii]

Umm.. lol
In a few months, jm's gonna come back to this thread and think to himself 'oh damn, I made such a fool out of myself'

 

[wolfhunter2]

The answer is:

.40 m for max height
& 5.80 ms-1 at 30 abv horizontal.

So you are right.

 

[clintmyster]

The answer is:

.40 m for max height
& 5.80 ms-1 at 30 abv horizontal.

So you are right.

the poor guy who posted mustve been confused by all that by jm haha

 

[lolokay]

wtf was the argument even about

 

[shady145]

wtf was the argument even about

jmo1 was trying to tell every1 that a projectile doesnt ston in the vertical component of motion, even if it has the property v(y)=0

Trebla and every1 else raped him and disagreed.