View Full Version : A physics question

I found this interesting and I thought that I might ask you to get your opinion. I know the answer and got it right on the test.

If a ball is thrown in a vertical direction what is happening to the ball with (with respect to acceleration)? Pretend that air resistance is negligible.

Chad

Plumbtex

02-09-2009, 09:27 PM

Depends, did you throw it up or down?

bucheron

02-09-2009, 09:35 PM

Depends, did you throw it up or down?

The question being with respect to acceleration, it is experiencing an acceleration downwards, so it does not matter which way it was thrown. I would go further, it doesn't even matter if it was thrown sideways it still experiences acceleration downwards.

was a physics teacher once. Hope I haven't disgraced the calling somehow, with this answer.

cheers

The ball is thrown upward, I meant to say that but didn't.

Norm has the idea. The velocity of ball thrown upwards is stated by V(original)+AT. Where V(original) is the initial velocity, A is acceleration and T is time.

So the effect of gravity on the ball is the acceleration of 9.8 m/s. So with that in mind if the ball is thrown upward which way is the ball accelerating? What happens when the ball's velocity is zero? If the ball's velocity is zero is it accelerating?

Chad

it's not accelerating when it leaves your hand.

Plumbtex

02-09-2009, 09:41 PM

If thrown upwards wouldn't it first need to decelerate before accelerating downwards or is a deceleration in one direction the same as an acceleration in the other?

Of course if you taught physics you know way more about this than I do. But from what I know you have it correct.

For example if you fire a bullet out of a perfectly horizontal gun and at the exact same instant you drop another bullet from the same height, both bullets will hit the ground at the same time. This is due to the fact that the acceleration from the effects of gravity acts on both bullets at the same time and they both accelerate toward the ground at the same time.

My question, which has not been answered yet, is does the curvature of the earth play into this equation?

Chad

acceleration a deceleration are the same thing.

acceleration is the change in velocity over time. Because velocity (http://en.wikipedia.org/wiki/Velocity) is a vector (http://en.wikipedia.org/wiki/Euclidean_vector), it can change in two ways: a change in magnitude and/or a change in direction.

Chad

Nanoose

02-09-2009, 09:45 PM

You DID say that, Chad... the other guys missed it.

Uh, no Norm, he actually didn't. You assumed it ;)

You DID say that, Chad... the other guys missed it.

Tsk, tsk, Norman. "Vertical" is not the same as "upwards".

When the ball's velocity is zero relative to a plane of reference, such as the earth, its acceleration is zero.

Velocity is always relative to a frame (not a plane) of reference.

Kaa

I've got to disagree with you on a couple of points. The formula is correct, although you are correct with the negative. The A (or acceleration) would be expressed as -9.8 m/s. The formula would be written V=V(0)+(-A)*T.

The ball even at the instant it leaves the hand is accelerating toward the earth. When the ball reaches the maximum height and the velocity is zero, the ball is still accelerating toward the earth at 9.8 m/s.

Chad

Nanoose

02-09-2009, 09:50 PM

What OTHER direction is 'vertical'? :)

Uh, duh...:rolleyes:

Keith Wilson

02-09-2009, 09:51 PM

If a ball is thrown in a vertical direction what is happening to the ball with (with respect to acceleration)? Pretend that air resistance is negligible.Exactly the same thing that happens if you drop it, or throw it horizontally; constant 9.8 m/sec^2 acceleration down (toward the center of mass of the earth, to be pedantic).

When the ball's velocity is zero relative to a plane of reference, such as the earth, its acceleration is zero.This is wrong. Acceleration is dv/dt, the rate of change of velocity over time. Zero velocity does not mean zero acceleration; you can have any value of acceleration in any direction at the point that velocity is zero.

Quit picking the fly**** out of the pepper.... it was clear that he meant upwards. What OTHER direction is 'vertical'? :)

Well, "vertical" is not a direction to start with :D

You're working WAYYYY too hard at this, Kaa! :o

Nah. Please notice that I didn't mention other planets, being underwater, and a variety of other stuff that make physics exercises fun :D

Kaa

Captain Blight

02-09-2009, 10:05 PM

and a variety of other stuff that make physics exercises fun :D

KaaI used to enjoy that stuff out behind the FFA shed in high school between 5th and 6th periods.

This is wrong. Acceleration is dv/dt, the rate of change of velocity over time. Zero velocity does not mean zero acceleration; you can have any value of acceleration in any direction at the point that velocity is zero.

This is what I was saying in my last post.

And yes I should have also said that we were standing on planet earth and not underwater. Although I did include the disclaimer about air resistance.

Chad

Side note: When uploading your straight line motion lab report to the dropbox on the server make sure that you upload the straight line motion lab report and not Assignment #2 from Construction Planning & Scheduling.

Now back to your regularly scheduled program.

Oh also be advised my Statics II instructor is the same as my Statics I instructor and he sux worse at Statics II than he did in Statics I.

Chad

George Roberts

02-10-2009, 01:05 AM

The problem clearly states to ignore air resistance. This leads me to believe that only air resistance should be ignored.

As the ball rises the gravitational acceleration decreases as the square of the distance between the centers of the earth and the ball increase. Anything else moving on the earth (or the rest of the cosmos) also affects the acceleration.

The reference frame needs some attention. A reference frame fixed in space will provide a rather complex but real value for the acceleration of the ball. Certainly don't make the error of fixing the reference frame to the earth without including the motion due to the earth rotating about its axis and revolving around the sun, the sun revolving through our galaxy, ...

As long as we are too weak to reach escape velocity we can include only the rotation of the earth about its axis and come up with a reasonably accurate general solution to the problem.

Once we reach escape velocity we get into very complex paths some of which are almost periodic in some frames of reference, some of which are chaotic. The resulting accelerations are perhaps best said to be time and location dependent.

Air density v. ball density affects acceleration. Since this is distinct from air resistance, a ball that has litte densitywill have an interesting acceleration future as buoyancy forces cause it to rise to the "surface" of the atmosphere. On the other hand a ball that is dense enough to penetrate the earth as it falls will have a different but equally interesting acceleration future.

I leave it to others to mention the effects of relativity.

On the other hand perhaps the ball is small enough that we should consider quantum effects ...

Anyone who would accept the simple answer, that the acceleration is constant, is expecting too little.

BarnacleGrim

02-10-2009, 01:11 AM

It's perfectly simple. From the point the ball leaves the hand it has a constant acceleration of -9.8 m/sē

Nothing else matters unless the ball is the size of a planet.

Captain Blight

02-10-2009, 02:09 AM

who are without the synapses and neurons that ,via electrical energy, coordinate all our movements and thoughts?

Well, Dutch, for one.

BrianW

02-10-2009, 04:56 AM

My dog caught the ball. :(

P.I. Stazzer-Newt

02-10-2009, 05:05 AM

Question 2:

The ball leaves the hand with a velocity of 11211 m/s - how long do you have to wait for its return?

Ignore air resistance if you like.

shamus

02-10-2009, 05:13 AM

2285.63 seconds - about 38 minutes, according to the formula above, except that it's probably gone into orbit.

shamus

02-10-2009, 05:29 AM

Although if that's the exact escape velocity for a vertical launch, I doubt if a ball will fly strait enough. If it gets up a bit of a spin, it will depend on whether it veers east or west. I think.

P.I. Stazzer-Newt

02-10-2009, 05:33 AM

Escape speed is a scalar.

shamus

02-10-2009, 05:47 AM

Yeah but,

From wiki

The escape velocity relative to the surface of a rotating body depends on direction in which the escaping body travels. For example, as the Earth's rotational velocity is 465 m/s at the equator, a rocket launched tangentially from the Earth's equator to the east requires an initial velocity of about 10.735 km/s relative to Earth to escape whereas a rocket launched tangentially from the Earth's equator to the west requires an initial velocity of about 11.665 km/s relative to Earth. The surface velocity decreases with the cosine (http://en.wikipedia.org/wiki/Trigonometric_function) of the geographic latitude, so space launch facilities are often located as close to the equator as feasible,

Is that relevant?

George Ray

02-10-2009, 05:51 AM

Check out Walter Lewin's MIT physics lectures, available free through Open University, also on iTunes.

http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/

Nicholas Scheuer

02-10-2009, 07:10 AM

"Decellerating" is just another way of expressing negative accelleration.

BTW, the path described by the ball through space is parabola, due to the fact that the earth will continue rotating. No, Wait! The ball will have had some lateral velocity when it left the hand, so that velocity must be lost, so the path will be a sort of "lopsided parabola".

Moby Nick

Tealsmith

02-10-2009, 07:45 AM

What goes up, must come down.

Popeye

02-10-2009, 08:02 AM

If a ball is thrown in a vertical direction what is happening to the ball ?

it interacts with the universe

huisjen

02-10-2009, 08:10 AM

http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/

Enjoy.

Dan

Popeye

02-10-2009, 08:13 AM

I'm not sure of the question. Newtonian physics is great stuff because it's directly observable. Einstein's relativity makes you sweat. It's an affront to common sense. Without saying it's wrong. I wouldn't so presume.

i think the warning is to not always trust in your own observations

It's perfectly simple. From the point the ball leaves the hand it has a constant acceleration of -9.8 m/sē

Nothing else matters unless the ball is the size of a planet.

That is correct as far as I know. Even when the ball is at the top of its flight and has zero velocity it is still accelerating at 9.8 m/s^2 toward the earth.

Chad

huisjen

02-10-2009, 11:02 AM

Momentum is not acceleration. Momentum is speed times mass or resistance to acceleration. Acceleration is force divided by mass.

Dan

huisjen

02-10-2009, 11:11 AM

The hand has accelerated upward.

As soon as it leaves the hand it's being accelerated downward by gravity.

When it leaves the hand, its downward speed is a negative number, but it's increasing. When it hits its peak, downward speed is zero, but it's increasing. As it starts down, it's downward speed is a low positive number, and increasing.

Dan

Keith Wilson

02-10-2009, 11:48 AM

Isn't the ball still speeding up when it first leaves the hand before it starts to slow down?No. What would make it speed up? It's just sailing through the air with nothing pushing on it. As soon as the ball leaves your hand, it starts to slow down. At the top of its trajectory, it stops momentarily, then starts to speed up again.

The acceleration (defined as change in velocity per unit time) is always downward. Since it starts off with an upward velocity, that decreases until the ball pauses at the top of the arc, then as it moves downward it speeds up - but acceleration is constant.

N

The acceleration (defined as change in velocity per unit time) is always downward. Since it starts off with an upward velocity, that decreases until the ball pauses at the top of the arc, then as it moves downward it speeds up - but acceleration is constant.

Exactly. So with that in mind that same ball is being subjected to an external force that is equivalent to the acceleration due to gravity.

Chad

George Roberts

02-10-2009, 03:03 PM

"velocity"

Velocity is a vector - has a magnitude and a direction, so in the "simple" gravity case the velocity always decreases.

Speed is the scalar - the magnitude of the velocity vector, that decreases as the ball rises and then increases as the ball falls.

It is important to use the correct term or a lot of the talk becomes confusing.

Rapelapente

02-10-2009, 05:28 PM

My question, which has not been answered yet, is does the curvature of the earth play into this equation?

Chad

Yes, not exactly the curvature of the earth itself, but your position from the equator. Since the earh is spinning, the ball is not only subject to gravitation acceleration, but also to Coriolis acceleration. So the trajectory will not be straight, but an ellipse, deflected to west on the way up and to east on the way down.

Excepted if the "thrower" is on the equator, where the Coriolis acceleration is null.

At 45° of lattitude, the deflection will be about 15millimeters for 100 meters of displacement, so you must be an Olympic champion to be able to see any alteration of the trajectory...:D

could be close to that:

http://farm2.static.flickr.com/1038/3269842973_199824f0b7_o.jpg

epoxyboy

02-10-2009, 06:19 PM

Now that piece of complex physics has been sorted to sorted to the satisfaction of all, design and build of that thermonuclear weapon should be a piece of cake.

Pete

George Roberts

02-10-2009, 07:11 PM

" Excepted if the "thrower" is on the equator, where the Coriolis acceleration is null."

I believe you intended to say at the poles.

bucheron

02-10-2009, 10:06 PM

My answers, speaking as a junior high school physics teacher would.

Q. . . if the ball is thrown upward which way is the ball accelerating? A. Downward.

Q. . . What happens when the ball's velocity is zero?

A. It changes direction, from upward to downward. This happens in an instant, that is, a location in time with no length in time. (Students are being asked to think abstractly here)

Q. . . If the ball's velocity is zero is it accelerating?

A Yes. It is accelerating at all times in this process.

Critique of statements.

Norman... it was clear that he meant upwards. What OTHER direction is 'vertical'?

A: downwards can be vertical.

Keith... down (toward the center of mass of the earth, to be pedantic).

A: At the scale usually assumed in these tests, objects are assumed to fall parallel to each other, rather than towards a single point.

Chad... the disclaimer about air resistance.

A. Teachers often throw in the phrase "in a vacuum.." This tells the students "Ignore air resistance."

George... A.the points raised are true, but could only be raised by a student who had moved beyond elementary physics. Having such a person in one's class is a damned nuisance.

Shamus...2285.63 seconds

A. My answer is 2288s. Finding and explaining these small answer differences to students is also a nuisance. I should now go into rounding and significant figures. Before calculators, G was usually given as 10 m/s^2 to avoid some of the difficulty.

P.I. ... Escape speed is a scalar.

A. It has direction (away from the planet) so it is a vector, and should more properly be called escape velocity.

Shamus re Wiki quote . . . .

There were no cues given in the original question to indicate these were relevant.

Nicholas.. "Decellerating" is just another way. . . A: Correct!

.. BTW, the path described by the ball through space . . .

A: you have introduced uncued irrelevancies. Saying "through space" is good. A teacher would avoid saying "thru the air" because this is cuing the students "you must take air resistance into account"

Huisjen . . Acceleration is force divided by mass.

A: true but way out of context. Memorize this definition "Acceleration is the time rate of change of velocity".

Chad ... an external force that is equivalent to the acceleration due to gravity.

A: the force is gravity, which causes the acceleration.

Chad, I see you are a Construction Project Manager. What test were you taking and how important to you was it?

I would like to see the question, verbatim as put to you, and the teachers standard answer.

In an elementary physics class, if the teacher says "A brick falls off a scaffold ..." the students are being cued to answer a question about gravity and its effects. As a construction project manager hearing that, my guess is that your thoughts would be

"Was anyone injured?"

"was anything important damaged?"

"How long will the project be held up?"

To the Forum

I have spent some hours composing these responses, and have enjoyed the opportunity to practise my rusty skills. I wish I could have given equal attention to all of you, because I know that some of you will vandalize your desks or harass other students in an attempt to make me notice you. I want you to know that I noticed you all and I care for you.

Really.

cheers and get your assignments in on time.

chrisk

02-11-2009, 12:22 AM

Ooh can I play???

The first thing you have to do is settle on a frame of reference. Some of the issues here is just that they picked a different frame of reference then others.

There are two postulates to Einsteins theory of relativity:

1. There is no preferred or absolute inertial system. That is, all inertial frames are equivalent for the description of all physical laws

2. The speed of light in a vacuum is the same for all observers who are in uniform, rectilinear, relative motion and is independent of the motion of the source. It's free space value is the universal constant c given by Maxwell's equations.

#1 is often referred to as the principle of relativity and people mistake that to mean that "Everything is relative". That's not what it means. It just means you can pick any darn place you want to measure the movement of anything else.

Therefore, I pick the ball's frame of reference and so it never accelerates nor does it move at all. When looking at things from this frame of reference your throwing the ball actually sent the earth hurtling downward and then back up to the ball when it eventually hits the ground. But, since you asked about the motion of the ball and not of the earth the answer is it didn't move at all, from the frame of reference that I chose.

#2 is a bit of a brain teaser if you think about it a minute. It means if I am traveling at you on a train moving at the half the speed of light and I shine a flashlight at you standing on the ground, both of us will measure the speed of light to be c. You'd expect that you'd see it at 1.5*c since the train I am on is moving at .5*c with respect to you and the light beam is traveling at c from me.

Anyway, in order for that thought experiment to workout you have to abandon the notion that distance and time are the same for all observers and non-intuitive things like space contraction and time dilation fall out of all of this. Also, it means addition of velocities is not described by simple arithmetic's addition process. I think there is a 1/(1-v**2/c**2)**1/2 factor in there somewhere, where v is the speed between the two frames of reference. Since this approaches 1 for velocities that are small compared to c you get the normal addition and normal Newtonian mechanics.

All of the "uniform, rectilinear" stuff in #2 means the frames of reference only have constant velocity differences, no acceleration or change of direction (another form of acceleration). So, this is just Einstien's Special theory of relativity. If we allow the different frames of reference to accelerate with respect to one another we move into Einstein's General theory of relativity. Even weirder yet and mathematics intensive.

chrisk

02-11-2009, 01:07 AM

So the trajectory will not be straight, but an ellipse, deflected to west on the way up and to east on the way down.

Excepted if the "thrower" is on the equator, where the Coriolis acceleration is null.If it were true that it deflected to the west on the way up and to the east on the way down wouldn't they cancel each other out and it'd land in the exact same place ?

It's been decades since I did these kinds of problems. I do agree that there would also be a horizontal component as well as a vertical component due to the rotation of the earth. I think it would work like this.

Let's pick a frame of reference to be on the surface of the earth where the origin is at the person standing on the earth's surface and the y axis is positive in the direction opposite the pull of gravity and the x axis is positive in the east direction.

As the ball left the hand, both the ball and the hand would have the same constant velocity toward the east due to the rotation of the earth. As the ball left the hand it would retain that eastward velocity, we're assuming no air friction, and would be identical to the hand for the duration of it's flight. As it gained altitude that same velocity would subtend a smaller arc compared to the hand traveling with the same constant velocity on the shorter radius with respect to the earth axis. So, as the ball goes up and back down the, hand travels a farther arc since all of it's time was spent on a shorter radius. So, even though both the hand and the ball travel with the same constant eastward velocity the ball will subtend a smaller arc spending it's time on a longer radius from the axis of rotation, but at the same eastward speed. So, the ball would land slightly to the west of where it was thrown.

I think it would always be deflecting to the west thorough out it's flight, more so at the top of it's flight and less at the bottom when it's altitude is similar to the hand's.

I also think that the horizontal deflection would be the greatest at the equator and non-existence if he was standing on the north or south pole, because the eastward velocity is greatest at the equator and there is no eastward velocity at the poles.

shamus

02-11-2009, 01:12 AM

Bucheron,

thank you for the effort you put in as far as I'm concerned, and I found some of your replies quite amusing.

Now I want to know all about centrifugal force...

George Roberts

02-11-2009, 01:21 AM

Bucheron ---

Tidy response. Next time you get to lead the class.

Rapelapente

02-11-2009, 03:20 AM

" Excepted if the "thrower" is on the equator, where the Coriolis acceleration is null."

I believe you intended to say at the poles.

SHAME ON ME ! you're right Georges & Chrisk! I inversed sinus and cosinus....:o

The Coriolis effect is null at the equator for horizontal speeds but not for vertical speeds.

Anyway it does'nt change anything at latitude 45° where I calculated my example.

But for Chrisk : No the two deviations DO NOT cancel each other, since the initial speeds are different: vertical on the way up and horizontal on the way down (only the vertical speed comes to zero on top, while gravitational acceleration is supposed constant). The ball falls WEST from the throwing point! The east component of coriolis acceleration during the way down is cancelling only the speed gained toward west, and not the distance done on the way up, so the speed will be again only vertical when landing.

I'll even dare to add that: the landing point will be a little bit south west, since the horizontal speed toward west will also creates a coriolis force to south...:)

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