everybody who took physics in high school knows the typical (and 99% correct) answer is "they will both hit the ground at the same time." In reality, its far more complex... and beyond most of us who dont study physics in particular. Although I seriously doubt that a bullet traveling at 200 mph will be going fast enough to be affected more than a trivial amount by the curvature of the earth.
yhelothar
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Originally posted by: soydios
Originally posted by: MrPickins
Both land at the same time.
Go ask Newton.
/agree
The horizontal components of each bullet's velocity vector doesn't matter. Only the vertical component matters. If the fired bullet is shot at perfect horizontal, both the fired and dropped bullets will hit the ground at the same time, because they have the same vertical velocity vectors.
Hooray for AP Physics. I am such a nerd.
Too bad AP Physics' simplified/perfect world isn't too helpful in solving these real world problems.
Hooray for college engineering physics!
mrSHEiK124
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- Mar 6, 2004
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DivideBYZero
Lifer
- May 18, 2001
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Originally posted by: soydios
Originally posted by: MrPickins
Both land at the same time.
Go ask Newton.
/agree
The horizontal components of each bullet's velocity vector doesn't matter. Only the vertical component matters. If the fired bullet is shot at perfect horizontal, both the fired and dropped bullets will hit the ground at the same time, because they have the same vertical velocity vectors.
Hooray for AP Physics. I am such a nerd.
The air resistance is a force that is applied directly against the direction of movement, and its intensity (value) increases with speed (with the squared speed I think, but depend on speed). As a result, the faster bullet will lose more of its vertical "falling" speed because of the increased total velocity.
(but I might be wrong)
Whitecloak
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Demon-Xanth
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Depends on the car. C&D drove a stock bodied Firebird to 220MPH at Bonneville and ended up with the car pointed to the sky and about 20 feet off the ground
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in the last bullet thread it was determined that while the curvature of the earth may be small, it still affects which bullet hits the ground first. it was the stationary dropped bullet, of course. so, while it may be an even more trivial amount, it will enter into the equation. if the extra 200 mph is not somehow negated then that bullet will hit later, ceteris paribus.
middlehead
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KillerCharlie, nice to see someone in the aerospace field. Awesome, we need you around here.
:beer: for your knowledge in this thread.
EDIT: Where do you work? (And, please say Lockheed Martin skunkworks, that would be like totally awesome and I would have to worship you).
:beer: for your knowledge in this thread.
EDIT: Where do you work? (And, please say Lockheed Martin skunkworks, that would be like totally awesome and I would have to worship you).
Originally posted by: KillerCharlie
Sort of. I'll try to explain it in words, though it'd be easier to explain mathematically.
Let's just say the instant they're dropped, they each fall at about 10 mph. I know this isn't true, but it's the same effect and it illustrates the point.
To find the drag, we have to find the total velocity on the bullets (actually the total velocity squared, since drag is largely a function of V^2). Remember velocity is a vector (has direction) so we square both components. I'll leave it in terms of V^2 since drag is mostly proportional to V^2.
Moving bullet: 200mph forward and 10mph down = 200^2 + 10^2 = 40100 (mph)^2.
Dropped bullet: 10mph down = 10^2 = 100 (mph)^2
Let's just use these numbers as drag, so moving bullet has 40100 units of drag and the dropped bullet has 100 (but in different directions). Now we have to find the component of the drag that is pushing upward on the bullet (the drag slowing its downward fall).
Moving bullet: Drag in vertical direction = 40100 * (sin(angle)) = 40100 * (10/sqrt(40100)) = 2002.5
Dropped bullet: Drag in vertical direction = 100
Therefore, the bullet shot from the moving car will drop more slowly. Its drag force in the vertical direction is generally higher than the bullet that was dropped.
Time for bed... I'm working in the wind tunnel tomorrow
But wouldn't any change in upward pressure by the higher velocity of the air particles striking the bottom of the bullet be evened out by the increased downward pressure smacking the top of the bullet, thereby canceling out the effect?
Assumming it's shot straight and there's equal area above and below it.
I'm not sure I understand what you're saying. Typically, the pressure is greater in the direction that faces the wind. You basically can never have the back pressure greater than the front pressure, although they can be close. If the front/back pressure are the same, there will STILL be drag, and quite a bit drag at that. This is called skin friction drag, and is caused by the air dragging along the object's surface.
Also, it is a fallacious idea that air particles "hit" the object. They only hit an object at one singular point near the front called the stagnation point. Thinking of air particles hitting an object and bouncing off is bad thinking. If a particle hits and bounces off, then where would the particle behind it go? Instead, the air particles flow around an object, always in a direction tangent to the surface.
I work for the other company... and in commercial aircraft
HomeBrewerDude
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- Jan 18, 2001
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