If you shot a bullet straight up in a vacuum, then it would fall back to the ground at a speed equal to the muzzle velocity, assuming gravity were still present.. But if it were shot on Earth or in any other atmosphere, it would return to the ground at a slightly lower speed due to wind resistance. The drag force will do negative work on the bullet along the entire path of its trajectory, causing it to lose some of its original energy.
(Actually, dudas_91 is correct about terminal velocity being significantly less than the muzzle velocity of a typical handgun. But I would say that I'm not so much "completely wrong" as "missing a small detail." Anyway, I don't think his explanation of terminal velocity is clear, so I'd like to attempt one myself, while acknowledging that it didn't occur to me until reading the below response.)
When an object falls through an atmosphere, two forces act on it: gravity and drag. Gravity always pulls downwards, but drag opposes motion in any direction. So when an object is falling, drag acts upwards. Gravity is constant near the surface of a planet, but drag is proportional to velocity. The more quickly an object is falling, the greater the upwards force of drag. At a certain critical velocity, the upwards drag force and the force of gravity will be equal in magnitude and opposite in direction, meaning that there is no net force and the object continues to fall at constant velocity, the terminal velocity. However, both forces (gravity and drag) do continue to act on the object at all times.
2007-08-05 15:56:08
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answer #1
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answered by DavidK93 7
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In short, TRUE.
After the bullet leaves the gun it will decelerate in speed (due to gravity) until the bullet comes to a full stop and begins to drop. On the way back down it will accelerate until it reaches the ground.
The speed at which the bullet was traveling when it left the gun will be exactly the same as the bullet's speed just before it hits the ground.
Assuming the resistance due to friction remains constant throughout the bullet's trip, the only force acting on the bullet is gravity which is constant.
2007-08-05 16:05:46
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answer #2
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answered by java2bobby 3
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If you shoot a bullet up in the air, it is true that it will not come down as fast as it went up.
Since it is shooting up in air, its motion is resisted by air, therefore the speed must be definitely less than its initial speed, even if it attains a terminal speed on its return path.
If the initial speed is greater than the escape velocity it will never return to earth.
If the initial speed is such that it equals the terminal speed that it attains on its return path, then naturally it will come down as fast as it went up.
2007-08-05 18:09:44
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answer #3
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answered by Pearlsawme 7
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Not true. Everything falling is exposed to drag (a fancy word for friction caused by air). The bullet will be slowed down by air. At a point called terminal velocity the bullet will no longer accelerate due to gravity. Terminal velocity is the highest maximum velocity and is when drag doesn't allow the object to accelerate because of gravity. Read about terminal velocity.
Everyone is completely wrong. In our atmosphere the terminal velocity of a bullet is about 145 mph.
2007-08-05 15:58:10
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answer #4
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answered by dudas_91 4
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well, i noticed you said "shoot a bullet up in the AIR" which means no to your question. the bullet will only reach its terminal velocity because of the air resistance which means the bullet wont come down nearly as fast as it would have been shot up.
2007-08-05 19:21:27
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answer #5
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answered by ehrich69 1
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Yes, it is true that is will come down just as fast as it went up. This is because the path of any projectile (neglecting air resistance or any external force other than gravity) travels in the shape of a parabola. And of course, one distinct property of a parabola is that it is symmetrical, meaning that it's path up will exactly mirror its path down.
2007-08-05 15:54:50
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answer #6
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answered by C-Wryte 4
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Not at all.
What goes up must come down (as long as it doesn't go up at a speed equal to or exceeding the gravitational escape velocity.) But there are two forces acting on the bullet as it goes up, and only one force acting on it as it goes down.
The opposing forces acting on it as it goes up are the force of the gunpowder charge pushing it up, and the force of gravity pulling it down. The speed it travels up will equal the upward push minus the downward pull. But the force of gravity is the one force acting on the bullet as it falls back down, though it is certainly true that, the bigger the powder charge pushing it up, the higher it will go (up to a point), and, because the force of gravity increases by the square of the distance of the falling object from the center of the earth, the bullet will accelerate on its downward journey up to a limit imposed by the mass of the earth.
Note: different planets or other bodies in space have higher "escape velocities" and different rates of falling for falling objects. But the mass of an object is the same no matter where it is. That is where the difference between mass and weight comes in. A man who weighs 150 pounds on earth will weigh only 30 pounds on the moon, but he is just as massive on either planet, and, if he hits you with his fists, they will hit with as much force and hurt you as bad on the moon as on the earth.
The mass of the planet also dictates the escape velocity. A bullet fired from a small asteroid might not fall back down, but might continue traveling in outer space 'til it hits something out there. A pickup traveling a hundred miles an hour on a small asteroid might escape the gravitational pull and fly out into space.
A rock thrown upwards on the moon would go higher than one thrown upwards on the earth and would accelerate more slowly on its downward path, but rest assured that, if it hit you in the head when it came down, it would hit you with just as much force as the same rock falling from the same distance on earth, because the force of collision depends on the mass of the planet, moon, or asteroid attracting it.
2007-08-05 16:42:53
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answer #7
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answered by John (Thurb) McVey 4
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In the real world, the answer is no. In a textbook exercise where you are told to neglect air friction, the answer is yes.
2007-08-05 17:15:41
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answer #8
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answered by Stan the Rocker 5
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The short answer is no. But for a detailed answer please review the link below that talks about Maximum Falling Velocity.
http://www.newton.dep.anl.gov/askasci/phy00/phy00800.htm
2007-08-05 16:02:49
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answer #9
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answered by Scootch 1
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UMMM I GUESS SO B/C GRAVITY WILL BE PULLING IT TOWARD THE EARTH BUT THEN AGAIN EVERYTHING HAS AN EQUAL AND OPPOSITE REACTION. SO I NOT SURE...
2007-08-05 15:54:35
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answer #10
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answered by Anonymous
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