physics

A ball thrown at an angle of 40 degrees with respect to the ground strikes a building 10m away at a height of 3m above the ground. The magnitude of the initial velocity is?

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  1. Vf^2 = Vo^2 + 2gd = 0,
    Vo^2 + 2*(-9.8)3 = 0,
    Vo^2 - 58.8 = 0,
    Vo^2 = 58.8,
    Vo(v) = 7.67m/s = Vertical component.

    Vo = 7.67 / sin40 = 11.92m/s @ 40 deg.

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