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![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 Feb 2008 16:16:49 IST
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all those who gave the AITS yesterday, can any1 of u explain me the question on projectile being launched with acceleration into free space(gravity free)?
why is it that the velocity will remain the same(as given in the solution)????
For others who didnt give it the question was:
a projectile is launched vertically from a tower with a velocity Vo , the height pf the tower is Yo..its acceleration at any instant is give by a(vector)=k[ v(vector) X A(vector). A is given as A(vector) = c/y(-k^)...y is the vertical coordinate of the particle.......
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hey, its not the velocity that remains same, it is the speed.
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11 Feb 2008 16:28:09 IST
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yea so velocity is changing....(i might hv overlooked speed)...but even so how cum the graph of KE is a straight line???
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11 Feb 2008 16:28:47 IST
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acceleratn is in plane perndicular to velocity,, hence constant
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11 Feb 2008 16:30:32 IST
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SIMPLE YAAR IF GOT THE QUESTION RIGHT
DIFFERENTIATING BOTH SIDES OF UR EQN. OF ACCELERATION U GET VELOCITY IN THE LEFT HAND SIDE DIFF. RHS THAT TURNS OUT TO BE A CONSTANT .
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11 Feb 2008 16:34:07 IST
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@ AYUSHITANDON
i think u r correct......
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this word is so small that it is a foolishness to hate anyone.
so, we love all. |
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11 Feb 2008 18:20:27 IST
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The question is not at all clear. What is y(-k)^??? What is asked in the question?
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Will nip in at times to solve problems :)
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11 Feb 2008 18:34:48 IST
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then where does the concept of maximum height come into play as its a gravity free space
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