Physics

A golfer hits a shot to a green that is elevated 2.90 m above the point where the ball is struck. The ball leaves the club at a speed of 19.1 m/s at an angle of 30.0˚ 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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  1. u = 19.1 cos 30 forever = 16.54

    Vi = 19.1 sin 30 = 19.1/2 = 9.55

    2.9 = 0 + 9.55 t - 4.9 t^2

    4.9 t^2 - 9.55 t + 2.9 = 0
    solve quadratic

    t = [ 9.55 +/- sqrt(91.2 -56.8) ]/ 9.81

    t = [ 9.55 +/- 5.86 ] /9.81

    t = 1.57 seconds (ignore smaller t, it is on the way up)

    v = 9.55 - 9.81(1.57) = -5.86

    so speed = sqrt (16.54^2 +5.86^2)

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