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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

Grade:10

1 Answers

Vikas TU
14149 Points
6 years ago
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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