What was the initial speed of the arrow?

An archer shoots an arrow horizontally at a target 15 m away. The arrow is aimed directly at the center of the target, but it hits 48 cm lower. What was the initial speed of the arrow? (Neglect air resistance.)

m/s

1 Answer

  • Find flight time:

    0.48 = (1/2) g t^2

    t= sqr(2*0.48/9.8) = 0.31298 s

    vx = x/t = 15/0.31298 = 47.926 m/s

    vy = gt = -9.80.31298 = -3.067 m/s

    Initial speed of the arrow:

    v = sqr(vx^2 + vy^2)

    v = sqr(47.926^2+( -3.067^2)) = 48.02 m /s

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