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Projectile Motion and Free Fall


jollyhawley

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I'm trying to get across to my students that if a projectile is thrown horizontally and the same kind of object is dropped from the same height and at the same time, both objects will hit the ground at the same time (neglecting air resistance). I had them put a washer on a ruler, and hit another washer off the table with the ruler- so that have one washer drop and the other one be hit at (hopefully) the same time. A few of them weren't convinced. Do you have any ideas for demonstrations or videos to watch to help prove this? Thanks.

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I'm trying to get across to my students that if a projectile is thrown horizontally and the same kind of object is dropped from the same height and at the same time, both objects will hit the ground at the same time (neglecting air resistance). I had them put a washer on a ruler, and hit another washer off the table with the ruler- so that have one washer drop and the other one be hit at (hopefully) the same time. A few of them weren't convinced. Do you have any ideas for demonstrations or videos to watch to help prove this? Thanks.

 

The Mythbusters actually just tested this(using guns!). You may be able to find the clip on YouTube.

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Here's a write up about it over at Dot Physics:

 

http://blog.dotphys.net/2009/10/mythbusters-bringing-on-the-physics-bullet-drop/

If you didn’t catch the latest MythBusters (yeah! new episodes), they did something straight from the physics textbooks. Just about every text has this example of shooting a bullet horizontally and dropping a bullet from the same height. The idea is that they should hit the ground at the same time. No one but the MythBusters could actually show this demo with a real gun.

 

The Physics

I am going to do some calculations, but I want to first write about the physics that accompanies this idea (and you can actually do it your self without the gun). What physics principle does this demo show? Well, it shows two things. First, ... <
>

 

 

Here's the vids:

 

kVS_sNYUs60

 

5FTd6x6m2XY

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  • 3 weeks later...
Actually, the dropped object should strike first because it has less distance to fall due to the curature of the earth.

 

It doesn't quite work like that. Gravity is uniform and the object falls at an equal rate regardless.

 

The Mythbusters demonstration proved the point very well.

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It doesn't quite work like that. Gravity is uniform and the object falls at an equal rate regardless.

 

The Mythbusters demonstration proved the point very well.

 

If you fired the bullet at a large enough speed it would orbit. The problem assumes you can treat the region as being flat.

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If you fired the bullet at a large enough speed it would orbit. The problem assumes you can treat the region as being flat.

 

That's what I meant since the original question made no reference to a curved surface at all. :)

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  • 2 weeks later...
It doesn't quite work like that. Gravity is uniform and the object falls at an equal rate regardless.

 

The Mythbusters demonstration proved the point very well.

 

Rate of fall is not the factor suggested in consideration of the curvature of the earth. The issue is the gain of altitude for the bullet as the earth's surface drops away from the bullet's trajectory.

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over the distances concerned, the earth is flat. if you really want to be pedantic enough to consider a few parts per 10 million error then go ahead and do it in your own time. but the error limits on a class room demonstration are far bigger and its just not worth considering when other factors play a far more significant role in producing errors

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  • 5 weeks later...
over the distances concerned, the earth is flat. if you really want to be pedantic enough to consider a few parts per 10 million error then go ahead and do it in your own time. but the error limits on a class room demonstration are far bigger and its just not worth considering when other factors play a far more significant role in producing errors

 

A horse can win a race by the nose. Is my immediate advisor part of a tag team? A line tangential to a curve meets it at one point. Since 1492, the world has been round. A plus or minus is hardly challanged for however large or small the value to which it applies. How mute must a humble superscientist be to escape stigma of pedanticism?

Edited by dalemiller
goof
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eh? i'm not claiming that the earth is flat. i'm claiming that on the scales considered, the bumpiness of the ground plays a far greater role than the earths curvature. especially if this is a classroom scale demonstration.

 

maybe i've missed the point of what you said, it was rather obfusticated do you mind reiterating in clearer language?

 

oh and we've known the earth is round since the 6th century BC. significantly before 1492.

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A horse can win a race by the nose. Is my immediate advisor part of a tag team? A line tangential to a curve meets it at one point. Since 1492, the world has been round. A plus or minus is hardly challanged for however large or small the value to which it applies. How mute must a humble superscientist be to escape stigma of pedanticism?

 

The point is that realistically, the curvature of the Earth is going to be far less than the irregularities in the floor itself, the degree to which you can control release times, etc. You can't be a superscientist until you learn about margin of error. But yeah, in a perfect experiment, on an infinitely smooth Earth, it wins by a nose.

 

But as long as we're being pedantic, the Earth has been known to be round for thousands of years, not just since 1492. And if it hadn't of been, nothing that happened in 1492 would have proved it to be so, unless you believe the Bahamas really are in India. :)

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A horse can win a race by the nose. Is my immediate advisor part of a tag team? A line tangential to a curve meets it at one point. Since 1492, the world has been round. A plus or minus is hardly challanged for however large or small the value to which it applies. How mute must a humble superscientist be to escape stigma of pedanticism?

 

Pedanticism?

You're all crappy scientists.

Everybody else gets negative points for not simply calculating the difference due to the curvature of the earth.

And dalemiller for... also not calculating the difference.

 

Guys, you can make a pretty good guess using nothing but Pythagoras. And wikipedia will give you an even better formula if you're interested.

 

Is this a science forum, or a discussion forum? Shame on you all, lazy people.

 

Circumference of the earth = (roughly) 40,000,000 m (radius is 6378 km)

Trajectory bullet Mythbusters = (roughly - I didn't watch the movie, movie is removed from Youtube) 500 m

 

That means that the bullet traveled 0.0000125 part of the curvature of the earth. In degrees (the earth being 360 degrees) that is 0.0045 degrees, or 0.000078 rad.

 

Now, we draw a triangle, with a 90 degree angle at the point where the gun was fired. One line starts horizontally (parallel to the surface of the earth), and one perpendicular to the surface of the earth, all the way to the center of the earth.

From the center of the earth, we draw a 2nd line, at an angle of 0.000078 rad with the one coming from the point where the gun was fired.

That 2nd line will intersect with the line which goes horizontally.

 

And it should be slightly longer than the one exactly perpendicular...

 

Using Pythagoras to solve this:

Length of non-perpendicular line = [math]\sqrt{6378000^2+500^2=6378000.02}[/math]

 

So, the answer according to my calculation is actually 2 cm over a 500m trajectory.

 

We should be able to measure the difference in how long it takes to drop... but only using accurate measurements.

 

And for those who say that I filled in the wrong numbers... try to find better ones - I bet that the outcome does not change much.

the only big change of course is if you have a shorter or longer trajectory of the bullet before it hits the ground.

 

p.s. this also shows that this is only relevant if you're able to measure the angle of the gun's barrel up to at least 0.0001 rad, or about 0.005 degrees. the curvature of the earth is quite irrelevant if you don't aim exactly parallel to the surface of the earth.

 

The point is that realistically, the curvature of the Earth is going to be far less than the irregularities in the floor itself, the degree to which you can control release times, etc. You can't be a superscientist until you learn about margin of error.

2 cm is not all that small of an irregularity... assuming that the trajectory was indeed about 500 m.

 

But it's a good point anyway. I wouldn't be surprised if the floor is actually completely straight - since it's a single building... which does make the curvature irrelevant. Not because it's such a small difference... but because of the architect who designed a flat floor.

Edited by CaptainPanic
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it was closer to 500ft the distance the bullet travelled. plugging that in, we get 2.18 millimeters

 

so assuming its launched from 1 meter thats 0.218% error. given that the tolerances of a floor are greater than that we can assume that we can then assume that the curvature is negligble.

 

also, given that this is most likely to be repeated in a classroom scale, if we use a 10m distance travelled(on the big side for a class room) we get 7.8 micro meters. at that scale specs of dust will give larger errors.

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over the distances concerned, the earth is flat. if you really want to be pedantic enough to consider a few parts per 10 million error then go ahead and do it in your own time. but the error limits on a class room demonstration are far bigger and its just not worth considering when other factors play a far more significant role in producing errors

 

Washer off a table maybe, but for a bullet and other experiments it can become a significant error.

 

I've seen hydrodymamic test tanks long enough that the curvature of the Earth had to be taken into account, both for this reason and also for coriolis "force", even though at relatively low speeds.

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So, the answer according to my calculation is actually 2 cm over a 500m trajectory.

 

Ha HA! You neglected to calculate the actual difference in the time of the fall due to the difference in the height! Which turns out to be about 5 milliseconds, given that the original height was about 1 meter.

 

And as long as we're being pedantic, the height difference calculation was given (albeit in English units) in the comments of the dot physics post, so there was no need to recreate them

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Pedanticism?

I wouldn't be surprised if the floor is actually completely straight - since it's a single building... which does make the curvature irrelevant. Not because it's such a small difference... but because of the architect who designed a flat floor.

 

The architect might not have ruled against levels, plumb bobs and poured concrete.

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  • 1 month later...
Pedanticism?

You're all crappy scientists.

Everybody else gets negative points for not simply calculating the difference due to the curvature of the earth.

And dalemiller for... also not calculating the difference.

 

...

 

But it's a good point anyway. I wouldn't be surprised if the floor is actually completely straight - since it's a single building... which does make the curvature irrelevant. Not because it's such a small difference... but because of the architect who designed a flat floor.

 

My omission of calculations was an escape from factoring in bullet's heading, mean velocity and latitude in order to determine net centrifugal force.

 

...

 

Again about fallacious indoors/outdoors issues: Architect be darned, the mason's bubble level steers his bricks around the bend of Earth, plumb lines throughout the construction diverge. Top floors are bigger than bottom floors.

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