Drop a bullet vs. Shoot a bullet

[meltdown75]

holy sh!t, fastest thread ever.

no gun obsessions in this forum!

 

[MrDudeMan]

Originally posted by: Bigsm00th
ok here it is. this is assuming no curvature of the earth, its in a vacuum, and there is no wind resistance. if you factor in the curvature of the earth, these results change and the dropped bullet hits first. i could do those calculations but at this point in time i am too lazy.

Text

metric is easier to use in calculations, but you get the idea.

i realize i am quoting myself, but i wanted to make sure people see this because it has the calculations and explanations inside

 

[91TTZ]

Originally posted by: rahvin

Originally posted by: 91TTZ
Would the unknowledgeable people PLEASE leave this thread? You are spouting complete nonsense.

You are asking a lot! Just take it for what it is, humor!

You're right, I'm sorry.

 

[LeiZaK]

Yes, the dropped bullet would hit the ground first under the condition of no air resistance and a perfect average curvature of the earth. But when air resistance is considered, factors such as drag and lift affecting the fired bullet become relevant. And what are the atmospheric conditions? I think it could really go either way in a real world scenario.

 

[timswim78]

NEED POLL!

 

[91TTZ]

Originally posted by: Bigsm00th

Originally posted by: Bigsm00th
ok here it is. this is assuming no curvature of the earth, its in a vacuum, and there is no wind resistance. if you factor in the curvature of the earth, these results change and the dropped bullet hits first. i could do those calculations but at this point in time i am too lazy.

Text

metric is easier to use in calculations, but you get the idea.

i realize i am quoting myself, but i wanted to make sure people see this because it has the calculations and explanations inside

Nice work.

I wonder why we got different times to hit the ground, though.

You got 4.518 seconds, I got 4.988 seconds from that calculator.

 

[91TTZ]

Originally posted by: timswim78
NEED POLL!

Math doesn't work by democracy.

 

[MrDudeMan]

Originally posted by: 91TTZ

Originally posted by: Bigsm00th

Originally posted by: Bigsm00th
ok here it is. this is assuming no curvature of the earth, its in a vacuum, and there is no wind resistance. if you factor in the curvature of the earth, these results change and the dropped bullet hits first. i could do those calculations but at this point in time i am too lazy.

Text

metric is easier to use in calculations, but you get the idea.

i realize i am quoting myself, but i wanted to make sure people see this because it has the calculations and explanations inside

Nice work.

I wonder why we got different times to hit the ground, though.

You got 4.518 seconds, I got 4.988 seconds from that calculator.

i used 1 meter. also, those are the real equations straight out of a physics book. i can guarantee my answers.

edit: i meant to say equations, not calculations. i did the calculations lol...this was my favorite part of physics, so i remember it quite well.

 

[Armitage]

Originally posted by: 91TTZ

Originally posted by: Bigsm00th

Originally posted by: Bigsm00th
ok here it is. this is assuming no curvature of the earth, its in a vacuum, and there is no wind resistance. if you factor in the curvature of the earth, these results change and the dropped bullet hits first. i could do those calculations but at this point in time i am too lazy.

Text

metric is easier to use in calculations, but you get the idea.

i realize i am quoting myself, but i wanted to make sure people see this because it has the calculations and explanations inside

Nice work.

I wonder why we got different times to hit the ground, though.

You got 4.518 seconds, I got 4.988 seconds from that calculator.

He used 1 meter - slightly less then the 4 feet you used?

 

Uh, yeah, all of you holier than thou, egotiscal boys insulting--rather than helping--people for daring to answer a question they may not have known answers to or for daring to ask questions and learn physics.... roll; No wonder too many people are afraid to learn physics!

Originally posted by: 91TTZ
I already answered the real world question. The dropped bullet would hit first. The curvature of the Earth is not much, but it isn't negligible, either.

In the real world, from a mathematical standpoint, the shot bullet would have to fall slightly farther due to the curvature of the Earth.

/thread.

Hey, I am sitting back, reading and learning as a lay person. But your statement here threw me off. Am I missing something because it appears contradictory to what you said earlier here:

"4. However, if you shot the bullet at 17,500 miles per hour, the bullet would never touch the ground, since its rate of descent will match the curvature of the Earth. It would always be "falling" and never hit the ground, since the ground it's vertically falling towards is constantly curving away from it, due to the horizontal speed."

Forgive me if I don't understand, but I would appreciate explanation or clarification, as the two appear inconsistent to me. Thanks in advance.

 

[MrDudeMan]

Originally posted by: Armitage

He used 1 meter - slightly less then the 4 feet you used?

yes i just check the calculations in my spreadsheet and 4 feet yields his answer. the only discrepancy here is our choice of distance.

 

[MrDudeMan]

Originally posted by: DearQT
Uh, yeah, all of you holier than thou, egotiscal boys insulting--rather than helping--people for daring to answer a question they may not have known answers to or for daring to ask questions and learn physics.... roll; No wonder too many people are afraid to learn physics!

Originally posted by: 91TTZ
I already answered the real world question. The dropped bullet would hit first. The curvature of the Earth is not much, but it isn't negligible, either.

In the real world, from a mathematical standpoint, the shot bullet would have to fall slightly farther due to the curvature of the Earth.

/thread.

Hey, I am sitting back, reading and learning as a lay person. But your statement here threw me off. Am I missing something because it appears contradictory to what you said earlier here:

"4. However, if you shot the bullet at 17,500 miles per hour, the bullet would never touch the ground, since its rate of descent will match the curvature of the Earth. It would always be "falling" and never hit the ground, since the ground it's vertically falling towards is constantly curving away from it, due to the horizontal speed."

Forgive me if I don't understand, but I would appreciate explanation or clarification, as the two appear inconsistent to me. Thanks in advance.

he is saying the pull from gravity wouldnt be enough to make the bullet hit the earth (in a vacuum). it would continue to go in a curved path around the earth since its horizontal velocity would be pushing it in a straight line, but the gravity would be pulling it closer. the 2 forces would "equalize" and put the bullet into orbit.

 

[vshah]

Originally posted by: Kyteland

Originally posted by: 91TTZ

Originally posted by: gigapet

Originally posted by: Kyteland
The dropped one. Since the earth is a sphere, and not flat, the fired bullet will actually have farther to fall than the dropped one.

Note that this assumes that it is fired in a vacuum.

gravity doesnt change because the earth is curved. four feet from the ground in LA is the ssame distance as it is in maine

No, he has a point there. While the gravity is the same, if you held a long, straight ruler out in front of you, level, the farther it extends, the higher it would be off the ground.

So the bullet would have to fall farther since the earth's surface is not flat.

Very good.

What I'm curious about is the affect from air resistance. That would be different for every shape of bullet. Is there any kind of shear that forces it up/down? A Jet can lift off the ground when moving fast enough and the flaps are in the right position. Can something similar happen to a bullet before air resistance slows it substantialy?

I honestly have no idea.

but you've also got to take into account that the force due to gravity is always pointing towards the center of the earth, NOT parallel to the bullet's (or ruler's) path.

 

[MrDudeMan]

DearQT, without knowing some of these words, this may be hard to take in...

the bullet has a tangential velocity to earth. it, by itself, has no component of velocity that makes it want to fall. the earth has gravity that is radial from every point on its surface. at every point in the bullets path, it is experiencing radial acceleration toward the center of the circle and tangential velocity away from the surface of the earth. at each instant, it is wanting to move in a straight line away from earth because of its tangential velocity, but the radial acceleration pulls it closer to the surface. it will never reach the surface, though, because as long as both forces remain constant, it will always want to be moving away and toward the earth at the same time. the resultant vector would be a circle around the surface of the earth, or what you may know as an orbit.

 

[akubi]

Originally posted by: Bigsm00th

i used 1 meter. also, those are the real equations straight out of a physics book. i can guarantee my answers.

well if you wanted to simplify the conditions and find the time it takes for anything on earth to fall, you only need t=sqrt(2x/g).

didn't need to draw that wallpaper of an essay to prove your point.
i call shens on your claim that you are a physics major :laugh:

 

[MrDudeMan]

Originally posted by: akubi

Originally posted by: Bigsm00th

i used 1 meter. also, those are the real equations straight out of a physics book. i can guarantee my answers.

well if you wanted to simplify the conditions and find the time it takes for anything on earth to fall, you only need t=sqrt(2x/g).

didn't need to draw that wallpaper of an essay to prove your point.
i call shens on your claim that you are a physics major :laugh:

:roll: the point of this thread was to explain to people without a physics background how and why the answer is what it is. im glad you feel secure enough to post an equation that any braindead monkey could take from a physics book, but that proves nothing. i wrote my "essay" to prove my work.

edit: oh, and i am a physics major, and also a double major in computer engineering if that helps you. im not sure why that would be shens, but call shens all you want. that doesnt change the fact that its true.

 

[Cerpin Taxt]

http://zebu.uoregon.edu/~imamura/121/oct13/orbit.html

 

[91TTZ]

Originally posted by: Bigsm00th

Originally posted by: DearQT
Uh, yeah, all of you holier than thou, egotiscal boys insulting--rather than helping--people for daring to answer a question they may not have known answers to or for daring to ask questions and learn physics.... roll; No wonder too many people are afraid to learn physics!

Originally posted by: 91TTZ
I already answered the real world question. The dropped bullet would hit first. The curvature of the Earth is not much, but it isn't negligible, either.

In the real world, from a mathematical standpoint, the shot bullet would have to fall slightly farther due to the curvature of the Earth.

/thread.

Hey, I am sitting back, reading and learning as a lay person. But your statement here threw me off. Am I missing something because it appears contradictory to what you said earlier here:

"4. However, if you shot the bullet at 17,500 miles per hour, the bullet would never touch the ground, since its rate of descent will match the curvature of the Earth. It would always be "falling" and never hit the ground, since the ground it's vertically falling towards is constantly curving away from it, due to the horizontal speed."

Forgive me if I don't understand, but I would appreciate explanation or clarification, as the two appear inconsistent to me. Thanks in advance.

he is saying the pull from gravity wouldnt be enough to make the bullet hit the earth (in a vacuum). it would continue to go in a curved path around the earth since its horizontal velocity would be pushing it in a straight line, but the gravity would be pulling it closer. the 2 forces would "equalize" and put the bullet into orbit.

Exactly.

 

[akubi]

Originally posted by: Bigsm00th

Originally posted by: akubi

Originally posted by: Bigsm00th

i used 1 meter. also, those are the real equations straight out of a physics book. i can guarantee my answers.

well if you wanted to simplify the conditions and find the time it takes for anything on earth to fall, you only need t=sqrt(2x/g).

didn't need to draw that wallpaper of an essay to prove your point.
i call shens on your claim that you are a physics major :laugh:

:roll: the point of this thread was to explain to people without a physics background how and why the answer is what it is. im glad you feel secure enough to post an equation that any braindead monkey could take from a physics book, but that proves nothing. i wrote my "essay" to prove my work.

edit: oh, and i am a physics major, and also a double major in computer engineering if that helps you. im not sure why that would be shens, but call shens all you want. that doesnt change the fact that its true.

oh? and when did you prove your first equation v_f^2=v_i^2..blahblah?

in fact, my reasoning is more fundamental than your convoluted approach.
any object with acceleration a moves (1/2)at^2 (basically derives from the definition of acceleration)
so plug in g for a and 1m for x and solve for t. what more is there to say? your equation derives from a set of these fundamental equations.

 

[vrbaba]

Originally posted by: Literati

Originally posted by: vrbaba
cant believe theres is 5 page discussion on it.
where are yall's high school physics memory?
so unexpected of AT....well, just proves the great diversity of people posting here

We were to busy having sex and being cool to remember this not relevant in the real working world crap... nerd!

sorry if a 2 liner discussion takes u 4 yrs of college and a 10 page AT post to understand.

 

[Connoisseur]

Originally posted by: RedRooster
From 4' high you drop a bullet(without casing) and at the same moment you fire a gun exactly parallel to and 4' from the ground. Which bullet would hit the ground first?

edit- ok, you walk out in the middle of a field on a normal day(not in space, not in a vaccum) and try this. standard .22 calibre rifle. and we can assume the field is perfectly flat, with no hills of dirt sticking up, you get the idea. or it could be all cemented, with hundreds of yards of cement(or however far a .22 will shoot) EXACTLY flat the whole way, not curving with the Earth's surface along the way.

Here's the solution: On a flat earth both of them would hit at the same time. On a curved earth, the shot bullet still has a y velocity becase there is no such thing as a straight parallel to a curved object. There's only a tangent. Thus, the shot bullet would hit the ground a bit later. Since the earth is so big for all practical purposes any point on the earth is virtually flat. so from our perspective they would both hit the earth at the same time. however, the fired bullet would hit the earth a few milli/micro seconds later. /end thread

P.S. I didn't read all of the threads

 

[Suspicious-Teach8788]

Originally posted by: Beast1284
Common sense would say the 4' dropped bullet, but I'm probably wrong.

How's that common sense? lol

 

[MrDudeMan]

Originally posted by: akubi

oh? and when did you prove your first equation v_f^2=v_i^2..blahblah?

in fact, my reasoning is more fundamental than your convoluted approach.
any object with acceleration a moves (1/2)at^2 (basically derives from the definition of acceleration)
so plug in g for a and 1m for x and solve for t. what more is there to say? your equation derives from a set of these fundamental equations.

you are asking me to prove equations derived by centuries of physicists. i will ask you in the same way to prove what you are saying so you can see how stupid it is.

again, you completely missed the point. what you just said is elementary to me but thats because it is my area of interest/study. others may not see it that way. you are being stubborn and trying to flex you e-muscles by "calling me out", even though you know my explanation is correct.

 

[chuckywang]

Originally posted by: vrbaba

Originally posted by: Literati

Originally posted by: vrbaba
cant believe theres is 5 page discussion on it.
where are yall's high school physics memory?
so unexpected of AT....well, just proves the great diversity of people posting here

We were to busy having sex and being cool to remember this not relevant in the real working world crap... nerd!

sorry if a 2 liner discussion takes u 4 yrs of college and a 10 page AT post to understand.

It's only 3 pages for me. Change your settings to 60 posts/page. You won't regret it.

 

[Suspicious-Teach8788]

How about you guys just go GOOGLE Kinematics Equations before you start arguing.

Better yet open a physics textbook. OR just friggn integrate the equations..