physics

find the angle of projection at the horizontal range is twice the maximum height of a projecyile

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  1. Range is x = vo^2 sin2θ/g
    Max Height is y = (vo Sinθ)^2/2g
    Set range = twice max height
    The vo^2 will cancel, solve for θ

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  2. find the angle of projection at which the horizontal range is twice the maximum teight of a projective.

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