The acceleration due to gravity! 9.8 meters per second squared! That is, from the top of its trajectory once its velocity = 0.
2006-08-12 22:22:23
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answer #1
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answered by TwilightWalker97 4
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Acceleration is defined as a transformation in speed. for this reason you desire 2 instant velocities to make certain an acceleration. although, in projectile problems acceleration consistently happens interior the Y direction and is comparable to the gravitational acceleration or 9.80 one m/s^2 or 32.2 feet/s^2.
2016-12-14 05:06:50
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answer #2
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answered by ? 3
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As long as the projectile is in the earth's gravitational pull, through out its path it is pulled down.
It is the reason that its velocity is reduced as it goes up and its velocity is increased as it comes down.
It is the reason that its velocity increases from zero to 9.8m/s in one second,from the position at which its instantaneous velocity was zero.
If there were no such acceleration it will be at rest for ever.
2006-08-13 00:42:19
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answer #3
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answered by Pearlsawme 7
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I believe washishi has it! At the very peak the object will have zero vertical acceleration, (up or down) and will have a decelerating horizontal component. The vectors of the two components cannot equal an acceleration.
2006-08-13 01:17:19
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answer #4
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answered by Plato X 2
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the accleration of a projectile at the topmost point .even though its velocity is 0 its accleration is vertically downward and 9.8m/s because at the top most point the vertical component of velocity becomes '-ve' i.e. it points to the center of the earth where as the horizontal component is in motion with constant velocity.
2006-08-12 22:22:27
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answer #5
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answered by josyula 2
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1 g or 32 ft per sec sq due to gravity
2006-08-12 22:28:15
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answer #6
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answered by Anonymous
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it is not accelerating since it is not in motion at that time when it is at max height
2006-08-13 00:03:18
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answer #7
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answered by washishi 1
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acceleration due to gravity
2006-08-12 22:32:35
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answer #8
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answered by Anonymous
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arent they all zero at the top?
2006-08-12 22:20:28
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answer #9
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answered by Anonymous
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