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

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Vikas TU · Answer

given, v = (2/5)^(0.5)*v’ at max. height the speed is, H = v’^2/2g or v’ = root(2gH) at half . of max. height => H/2 = 2gHsin^2thetha/2g solve for thetha.

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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

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

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