during the throwing motion the ball is accelerated in the opposite dirrection as gravity. once the ball leaves the hand of the thrower however, the acceleartion is constant, and is equal to that of gravity, g = 9.80 m/s^2
2006-11-08 15:57:25
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answer #1
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answered by Anonymous
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I assume the ball is thrown up in earth's gravitational field.
Simple answer: The acceleration of the ball is a constant = -g
Note: The above is not with out an assumption: The gravitational field is treated to be uniform, in the range of distance of the ball's flight. However one can argue that the ball could be thrown up 100's of kms high. Then the fact that the acceleration due to gravity decreases with distance from the earth, should be taken in to consideration. The acceleration due to gravity at a height, h is given by: g(h) = g(h=0)x[1 -(h/R)], where R = radius of the earth.
For the long range such as above scenario, acceleration decreases as the ball ascends; and increases when it descends.
2006-11-08 16:15:12
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answer #2
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answered by Inquirer 2
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All of them, but not necessarily in that order:
Going up; launch - increase from zero, then decrease as gravity kicks in. It freezes at the top then slowly accelerates down either continuously or reaching a constant speed of descent, depending on how high it was thrown.
2006-11-08 15:59:47
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answer #3
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answered by Bart S 7
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I know I've answered that gravity is a constant. This is simplifying it. Actually, the acceleration is indirectly proportional to the distance squared. What this means is that acceleration gets less and less as you move farther from the earth. However, the change in acceleration is negligible when throwing balls in the air.
2006-11-08 15:57:09
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answer #4
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answered by something 3
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Initial acceleration will decrease until it reaches a point of 0. Then, the ball will start falling again. (Accelerating in the other direction or, put in other terms, acquire a negative acceleration).
2006-11-08 15:55:42
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answer #5
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answered by Anonymous
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F = M a
Once the ball leaves your hand as you throw it, there is no force acting upon the ball except gravity. Within the confines of a hand thrown ball trajectory, the force of gravity is constant; the mass of the ball is constant; therefore, the accelleration of the ball is constant.
2006-11-08 16:55:25
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answer #6
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answered by Holden 5
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Acelaration is 0 so it increases when you through it up and slowly decreases. Then as it falls back to earth it increases again.
Accelaration is only constant when something is moving at the same speed. i.e not moving or travelling at 100km/h (mph)
2006-11-08 15:57:48
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answer #7
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answered by Anonymous
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decrease
2006-11-08 16:06:45
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answer #8
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answered by Michael G 1
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it remains constant.
2006-11-08 15:54:25
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answer #9
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answered by mighty_power7 7
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