A golf ball is struck with a five iron on level ground. It lands 92.7 m away 4.40 s later. What was (a) the direction and (b) the magnitude of the initial velocity?

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  1. a. Xo*t = Dx = 92.7 m.
    Vo*4.40 = 92.7,
    Xo = 92.7 / 4.40 = 21.07 m/s. = hor component of initial velocity.
    Yf = Yo + gt
    Yo = Yf - gt,
    Yo = 0 - (-9.8)(4.4/2) = 21.56 m/s. =
    ver. component of initial velocity.

    tanA = Yo/Xo = 21.56 / 21.07 = 1.02326.
    A = 45.7 Deg. = Direction of initial velocity.

    b. Vo=Xo/cosA=21.07/cos45.7=30.15m/s =
    Magnitude of initial velocity.

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