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A golf ball is hit with an initial velocity of 50 m/s at an angle of 45° above the horizontal. How far will the ball travel horizontally before it hits the ground?

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2 answers

  1. Vo = (50m/s,45deg.).

    Xo = hor. = 50cos45 = 35.36m/s.

    Yo = ver. = 50sin45 = 35.36m/s.

    t(up) = (Vf - Yo) / g,
    t(up) = (0 - 35.36) / -9.8 = 3.61s.

    t(dn) = t(up) = 3.61s.

    T = t(up) + t(dn)=3.61 + 3.61=7.22s. =
    Time in flight.

    Dh = Xo * T = 35.36 * 7.22 = 255m. =
    Hor. distance.

    Alternate Method:

    Dh = Vo^2 * sin2A / g.
    A = 45 deg.
    Vo = 50m/s.

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  2. hyt

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