physics

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A boy standing in a ditch throws a baseball upward toward his father. The ball leaves his hand at ground level, with an initial speed of 15.0 m/s, at an angle of theta = 53.0 degrees from the horizontal. The boy's father reaches up and catches the ball over his head, at a height of 2.0 m above the ground. The father catches the ball on its way down. Calculate how long the ball is in the air. ( g = 9.81 m/s2)

  • physics -

    Vo = (15m/ws,53deg.).
    Xo = 15cos53 = 9.0mm/s.
    Yo = 15sin53 = 12.0m/s.

    t(up) = (Yf - Yo) / g,
    t(up) = (0 - 12) / -9.81) = 1.22s.

    Yf^2 = Yo^2 + 2gh,
    h = (Yf^2 - Yo^2) / 2g,
    hmax = (0 - (12)^2) / -19.62 = 7.34m.

    h = Yo*t + 0.5g*t^2 = (7.34 - 2)m.
    0 + 4.9t^2 = 5.34,
    t^2 = 1.09,
    t(dn) = 1.04s.

    T = t(up) + t(dn) = 1.22 + 1.04 = 2.26s
    = Time in air.

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