Wow, some amazing answers.
On the one hand it's admirable that some people want to take into account things like the curvature of the earth and aerodynamics.
On the other hand, if they want to be so detailed, why don't they take into account the timing of the shot being exactly simultaneous with the dropping of the bullet. And take into account just how parallel the rifle is to the ground, and take into account the terrain of the ground (not perfectly curved).
What's really important here is the spirit of the question: the principle of kinematics: that gravity acts on a dropped bullet the same way as it acts on the bullet shot horizontally. If one wants to look into minor details, then there are many to look at -- not just curvature of the earth and aerodynamics.
The questioner, I believe, is looking at an idealised situation: muzzle velocity parallel to the ground, t=0 the same for both bullets, no wind or other aerodynamics involved.
But if we're talking about anything other than ideal conditions, then truly, the results could go either way. And I do really mean either way.
PS: Just to satisfy myself, I calculated to see how much the curvature of the earth affects the trajectory of the shot bullet.
Assuming the bullets were dropped and shot from a height of 1.5m, and the muzzle velocity of the shot bullet is 850m/s (about 2790ft/s), it takes the dropped bullet 0.553s to reach the ground.
Now, in 0.553 seconds, the shot bullet travels only 470m (assuming a constant speed of 850m/s). Does anyone think that the curvature of the earth makes much of a difference 470m away? I calculated the "drop" to be 17mm (yes, that's millimetres). That will account for an extra 3 milliseconds for the shot bullet to reach the ground after the dropped bullet. But in a real-life scenario, the shot bullet won't even go that far because of air resistance.
PPS: I used 6,378,135m to 6,378,200m for the radius of the earth. Either figure gave the same result (to the thousandth of a millimetre) because the distance travelled by the bullet is so small compared to the radius of the earth.
2007-08-04 20:43:42
·
answer #1
·
answered by Anonymous
·
2⤊
0⤋
Both fall at the same rate and impact at the same time. A counter argument says the fired bullet is not parallel to the earth's surface but really tangential and is sometimes used to "prove" that the fired bullet must drop further because the ground curves and is not truly flat. The argument is false, however, as the fired bullet remains in the same gravitational field which is radial from the earth's center.
As proof? A rocket traveling at 17,500 MPH remains in orbit, falling along the gravitational field of the earth, and does not drift off into space.
A fun experiment is to shoot a ball and mark its impact point on the ground. Set a can at the firing height of the ball above the impact point. Shoot the ball again at the exact time the can is dropped. The ball will fall into the can.
2007-08-05 06:04:08
·
answer #2
·
answered by john l 1
·
0⤊
1⤋
They will hit the ground at the same time, assuming the fired bullet doesn't strike something in its path before the ground. The opull of gravity is equal on both bullets.
2007-08-04 17:38:20
·
answer #3
·
answered by Sammy Da Bull 3
·
0⤊
0⤋
In theory they hit the ground at the same time. In practice the bullet will be spinning from the grooves in the barrel. This may induce a small amount of lift, keeping it aloft very slightly longer.
edit: In my job I do work in which milliseconds sometimes matter. Precision can be everything. I did forget to account for the curvature of the earth:)
As far as I know, the only reason ballistically the bullet rises is to account for the difference in height of the bore and the sights. This way the shot should be dead on at 25 yards and again, due to gravity, at 100 yards, varying slightly with the muzzle velocity of the weapon.
2007-08-04 17:44:50
·
answer #4
·
answered by balloon buster 6
·
0⤊
2⤋
howdy. i think of i'm meant to respond to "the two bullets each and every the floor on the comparable time". yet i think of because of the curvature of the earth the bullet dropped from relax will hit first - because of the fact the bullet flying in the process the air will rather earnings in top via a tiny volume because of the fact the earth "falls away". Does everybody believe me?
2016-10-14 00:27:24
·
answer #5
·
answered by joleen 4
·
0⤊
0⤋
They hit the ground at the same time. Gravity is the same for both. This does not take into account curvature of the earth which means the one fired from the gun actuually falls further and therfore does take longer to fall.
2007-08-04 17:39:37
·
answer #6
·
answered by Charles C 7
·
0⤊
0⤋
If you fire the gun on the level, and the ground is level, both bullets will hit the ground at the same time.
2007-08-04 17:40:49
·
answer #7
·
answered by TitoBob 7
·
0⤊
0⤋
The law of gravity states that they will hit the ground at the same time. The differnce being that the one shot form the gun will have more force behind it
2007-08-04 17:40:38
·
answer #8
·
answered by Anonymous
·
0⤊
0⤋
They both hit at the same time. Gravity affects things the same way with no relation to velocity in other directions.
2007-08-04 17:39:02
·
answer #9
·
answered by theboman@sbcglobal.net 2
·
0⤊
0⤋
physics will tell you the bullets will drop at the same rate which is true but...anyone familiar whith ballistics knows a fired bullet will rise slightly on its flight path.therefore it has further to drop and the dropped bullet will hit first
2007-08-04 17:42:37
·
answer #10
·
answered by ? 5
·
1⤊
2⤋