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Published by dellcom 02-08-2008
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#1
By
Jay-qu
on
02-08-2008
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| Re: physics question You are right to assume this is a constant acceleration question. In these problems there are 5 variable - acceleration, starting velocity, final velocity, displacement and time. Normally it works out that you are given three of these and you have to find the forth. Your infomation can be paraphrased as "it takes an object 1.5 seconds to fall 30m under acceleration of 9.8m/s/s" This info can be used to find either the velocity at a hieght of 30m (  ) or when its about to hit the ground (useless..)From there you now have another set of infomation. An object is dropped from rest, accelerates at 9.8m/s/s and has final velocity of v (from above). These 3 bits of infomation can be then used to find a forth bit (perhaps the distance it fell  )I like to help people solve the problems rather than just giving the answer, so let us know how you go with that and if you need more help then feel free to ask J |
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#2
By
dellcom
on
02-08-2008
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| Re: physics question ok so here is what i came up with.. The Initial Velocity when the object is at 30 m = (Xf - Xi - .5(a)t^2) / t = (0 - 30m -.5(9.8m/s/s)(1.5s)^2)/(1.5s) Vxi = -12.7 m/s (over the interval of 30m - 0m) then Vxf = -12.7 m/s + (-9.8 m/s/s) * 1.5s Vxf = -27.35 m/s now i can use that Vxi as the Vxf velocity between the start height and 30m to find the original position. Xi = - (Vxf^2 - Vxi^2)/2a + Xf Xi = (-27.35^2 m/s - 0^2 m/s)/2(9.8m/s/s) + 30m Xi = 38.2 m So the final answer should be the object fell from a height of 38.2 meters? |
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#3
By
Jay-qu
on
02-08-2008
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| Re: physics question Quote:
Except that you may want to quote your final answer to 2 significant figures since that is the allowed accuracy from your initial information. J | |
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#4
By
dellcom
on
02-08-2008
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| Re: physics question Quote:
hmm the question gives 3 significant figures, are you referring to me using 9.8m/s/s for gravity for the 2 significant figures? | |
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