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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Sep 2007 21:40:22 IST
Accepted Answer [?]
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speed of ball 1 = 120 m/s
height reached ( upto where we consider) = 100m
time taken = t sec
speed of ball 2 = x m/s
height reached = 100m
time taken = t-3 sec
s = ut + at^2 / 2
ut + at^2 / 2 = ut + at^2 / 2
120 t - 5 t^2 = x ( t - 3) - 5 (t-3)^2
120 t - 5t^2 = xt - 3x - 5[ t^2 + 9 - 6t]
120t - 5t^2 = xt - 3x - rt^2 - 45 + 30t
90 t - 5t^2 + rt^2 + 45 - xt + 3x = 0
also
100 = 120t - 5t^2
t = 12 + root(124) = 12 + 2root(31)
now put value of t in the previous eqn. just solve ahead ( only calculations now) to get the answers.
*PLZ RATE*
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---------------------------------------------------------------
* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
* Your friendly neighborhood spideyunlimited |
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