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A golfer hits a shot to a green that is elevated 2.60 m above the point where the ball is struck. The ball leaves the club at a speed of 16.7 m/s at an angle of 33.0˚ 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.

  • physics -

    Vo = 16.7m/s[33o].
    Yo = 16.7*sin33 = 9.1 m/s.

    Y^2 = Yo^2 + 2g*h = 0
    hmax= Yo^2/2g = -(9.1)^2/-19.6 = 4.23 m.

    V^2 = Vo^2 + 2g*d
    V^2 = 0 + 19.6*(4.23-2.6)
    V^2 = 19.6 * 1.625 = 31.85
    V = 5.64 m/s.

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