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The speed of a projectile when it is at its greatest height is (2/5)raised to the power 1/2 times its speed at half the maximum height.The angle of projection is? Detailed answer I want
given,v = (2/5)^(0.5)*v’at max. height the speed is,H = v’^2/2gorv’ = root(2gH)at half . of max. height =>H/2 = 2gHsin^2thetha/2gsolve for thetha.
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