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![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 Apr 2008 13:44:12 IST
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what is d radius of curvature of d parabola traced out by d projectile at a point where d particle velocity makes an angle  /2 with d horizantal
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its time fr u to achive d goal wake up donot hesitate to do hard work |
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11 Apr 2008 13:45:53 IST
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the answer is v2 /g cos /2
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11 Apr 2008 13:48:52 IST
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at the given instant, the component of g perpendicular to the velocity provides centripetal acceleration for the curvature
g costheta/2 = v^2 / R
R = radius of curvature
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11 Apr 2008 14:03:18 IST
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na d ans is [u^2Cos^2(teta)] /[gCos^3(teta/2)]
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its time fr u to achive d goal wake up donot hesitate to do hard work |
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11 Apr 2008 14:05:05 IST
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Wat a subject
so do u want all the persons( who didnt give solution ) to be dead?
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SHREYA |
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