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A body is projected such that the horizontal range is twice the maximum height of projection. Find the angle of projection; and the time of flight if the velocity of projection is 5m/s

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Charles

  1. vertical ... 2[5 sin(θ)] / g = t
    ... h = 5/2 sin(θ) * t/2

    horizontal ... d = 5 cos(θ) t

    2h = d
    5/2 sin(θ) t = 5 cos(θ) t
    dividing by cos(θ) t
    and multiplying by 2/5
    tan(θ) = 2

    substitute back to find t

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    Scott

  2. I don't really get you scott

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    Charles

  4. Am I to substitute it this way 2[5(sin2)]/9.8.? Or will I make use of tan instead, since tan=2

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    Charles

  5. Do tan-1 to find theta (how do you guys make a theta on this thing). Sub that back into the very first eq to find t

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    Chanz

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