physics

A golfer hits a ball off a tee from an elevated platform 13m above the ground. if the golf ball leaves the tee at 85 m/s @ 15 degrees, find the flight time, the maximum height above the ground and the horizontal displacement of the golf ball.

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asked by Marina
  1. Vo = 85m/s[15o]
    Xo = 85*cos15 = 82.1 m/s.
    Yo = 85*sin15 = 22 m/s.

    Y = Yo + g8*Tr = 0 @ max ht.
    22 - 9.8Tr = 0
    9.8Tr = 22
    Tr = 2.24 s. = Rise time.

    0.5g*Tf^2 = h max = 37.7 m.
    4.9Tf^2 = 37.7
    Tf^2 = 7.69
    Tf = 2.77 s. = Fall time.

    T = Tr + Tf = 2.24 + 2.77 = 5 s. =
    Flight time.

    h max = ho + (Y^2-Yo^2)/2g
    = 13 + (0-22^2)/-19.6 = 37.7 m.

    Dx = Xo*T = 82.1m/s * 5s = 412 m.

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    posted by Henry

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