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March 26, 2017

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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 16.1 m/s at an angle of 41.1° 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 - ,

    original component of velocity up = Vi
    = 16.1 sin 41.1 = 10.6 m/s

    H = Hi + Vi t - (1/2) 9.81 t^2
    3 = 0 + 10.6 t - 4.9 t^2
    so
    4.9 t^2 - 10.6 t + 3 = 0
    t^2 - 2.16 t + .612 = 0
    t = [ 2.16 +/- sqrt(4.67-2.44) ]/2
    t = 1.08 +/- .75
    t = 1.83 seconds or .33 seconds
    Use the 1.83 seconds. The .33 is on the way up just after you hit it.
    What is the speed at t = 1.83
    horizontal speed is 16.1 cos 41.1
    = 12.1 m/s
    vertical speed:
    v = Vi -9.81 t
    v = 10.6 - 9.81(1.83)
    v = -7.35 m/s
    total speed = sqrt (12.1^2 + 7.35^2)
    = 14.15 m/s

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