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4 January, 14:46

A dart is thrown at some speed v at 45° above the horizontal, with the aim of hitting the bull's eye of a dartboard located 2.5 m away. if the dart leaves the hand of the player at a height located 20 cm below the bull's eye, at what speed v must the dart be thrown?

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  1. 4 January, 15:01
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    let us say that t = time so that for horizontal motion:

    t = distance / velocity * cos angle

    t = 2.5 / v cos 45

    while for the vertical motion, (s = height = 20 cm = 0.20m, t = time, g = gravity)

    s = ut - 0.5 gt^2

    0.2 = v sin 45 * 2.5 / v cos 45 - 0.5 * 9.81 * [2.5 / v cos 45] ^2

    0.2 = 2.5 tan 45 - 61.31 / v^2

    61.31 / v^2 = 2.5 - 0.2 = 2.3

    v = 5.16 m/s
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