physics (clear answer)

A golfer hits a shot to a green that is elevated 3 m above the point where the ball is struck. The ball leaves the club at a speed of 16.0 m/s at an angle of 38.5° above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.

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asked by jason

  1. vertical problem first
    Vi = 16 sin 38.5 = 9.96 m/s
    v = Vi - g t = 9.96 - 9.81 t
    h = Hi + Vi t - (1/2) g t^2
    3 = 0 + 9.96 t - 4.9 t^2
    4.9 t^2 - 9.96 t + 3 = 0
    t = [ 9.96 +/- sqrt(99.2-58.8) ]/9.81
    t = [9.96 +/- 6.36 ] /9.81
    t = 1.66 seconds in air
    v = 9.96 - 9.81(1.66) = -6.36 m/s when it hits
    U = 16 cos 38.5 = 12.5
    speed = sqrt(12.5^2 + 6.36^2)
    = 14 m/s

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    posted by Damon

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