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bullet/gravity exercise; is it true?

 
 
lmoberg
 
Reply Tue 1 Nov, 2005 08:28 pm
I recall a science exercise where it was said that if you drop a bullet from outside the barrel of a rifle at the same time as you pull the trigger, the fired bullet and the droped bullet will hit the ground at the same time. Of course this requires the fired bullet to be parallel with the ground, the ground to be completely level, and no obstructions. What I want to know is the proof of this premise. Any other conditions? Does it have to be in a vaccuum? Does the rifle velocity have anything to do with it? Does the bullet size or shape? We know the fired bullet is subject to gravity as well as the droped bullet. What else?
Thank you
Lmoberg@crosslake.net
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stuh505
 
  1  
Reply Tue 1 Nov, 2005 09:27 pm
Quote:

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I recall a science exercise where it was said that if you drop a bullet from outside the barrel of a rifle at the same time as you pull the trigger, the fired bullet and the droped bullet will hit the ground at the same time.


Yup, that's right.

Quote:
What I want to know is the proof of this premise. Any other conditions? Does it have to be in a vaccuum? Does the rifle velocity have anything to do with it? Does the bullet size or shape? We know the fired bullet is subject to gravity as well as the droped bullet. What else?


Objects only CHANGE their motion in response to forces that are applied to them. Period.

What forces are applied to a bull when you drop it? Just one force: gravity, pulling it down. This speeds up descent downwards from an initial speed of 0.

What forces are applied to a bullet that has been fired from a gun? Gravity, pulling it downwards. There is NO force in the direction that it is being shot. There was a quick initial impulse force from the explosion, which initially sped the bullet up from 0 in the horizontal to 900-3000 FPS in the horizontal...but after than first split second, there is no more force.

Technically, there are some forces pushing in the opposite direction of the bullet speed...basically drag force, which slows down the bullet.

In the downward direction, there is still only one force: gravity. It pulls the bullet down towards earth at the same speed as if it weren't moving horizontally.

Other conditions? Assume a spherical bullet, assume no drag...
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g day
 
  1  
Reply Sat 5 Nov, 2005 08:15 pm
Er this is simply not correct, or rather its an impossible set of pre-conditions on say the Earth! You can't have a perfectly flat surface on a globe, and on anything other than a globe you will have varying gravitational strength over the surface level!

The bullet must fall further if it is on a globe like the Earth.

Think this way, for every mile you travel the Earth drops 4 inches (helps being a surveyors son).

A very high powered rifle might have an initial muzzle velocity of over a mile a second. If you where 5 metres above sea level and did this experiment the dropped bullet would take exactly 1 second to hit water, but the fired bullet would travel 5.1 metres vertically so it takes 1.01 seconds to hit sea level.

On a perfectly flat surface your question works, but not on a spheroid!

Further if the bullet where travelling much above 25km/sec it would simply attain escape velocity and never hit the Earth's surface.

So the approximation is interesting but wrong. A Naval destroyer with a radar guided 8" canon can hit a target at 30 miles away (48 KM) when its barrel is elevated to 45 degrees - so its target is well over the horizion. Ignoring air resistance you can see its muzzle velocity is over 2,000 m/sec. Imagine the canon height is only 30 metres maximum. So a dropped shell takes around 2.5 seconds to hit water, but a shell fired level to the surface travels 5,000 metres, by which time sea level due to the Earth's curvature has dropped a over a foot!
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DrewDad
 
  1  
Reply Sat 5 Nov, 2005 09:15 pm
This is obviously not intended as a real-world scenario.
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rosborne979
 
  1  
Reply Sun 6 Nov, 2005 09:54 am
Re: bullet/gravity exercise; is it true?
lmoberg wrote:
I recall a science exercise where it was said that if you drop a bullet from outside the barrel of a rifle at the same time as you pull the trigger, the fired bullet and the droped bullet will hit the ground at the same time. Of course this requires the fired bullet to be parallel with the ground, the ground to be completely level, and no obstructions. What I want to know is the proof of this premise. Any other conditions? Does it have to be in a vaccuum? Does the rifle velocity have anything to do with it? Does the bullet size or shape? We know the fired bullet is subject to gravity as well as the droped bullet. What else?
Thank you
Lmoberg@crosslake.net


Hi LMO,

The answer to the question is YES, *if* you are working with a simple theoretical model (not the real world).

Usually the point of such a physics question is to get the student to understand that gravity affects object equally, whether they are moving or not.

Even in the real world, the actual results are very close to the theoretical, assuming you try hard to minimize variables such as tragectory, friction and "flatness" of the ground. If you stand in Death Valley on a calm day and fire a bullet from a gun which is aimed parallel to the ground, and drop a bullet from the same height at the same time, then both bullets will hit the ground at almost the same time (but at very different places).

But as others have mentioned here, there are many external conditions in the real world which will affect the actual results.
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