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An artillery shell is fired at an angle of 48.2 degrees above the horizontal ground with an initial speed of 1530 m/s. THe acceleration of gravity is 9.8 m/s^2. Find the total time and flight of the shell,neglecting air resistance.Answer in units of minutes.

Find its horizontal range,neglecting air resistance. Answer in units of km

  • physics -

    Vo = 1530m/s @ 48.2 deg.

    Vo(h) = 1530cos48.2 = 1020m/s.

    Vo(v) = 1530sin48.2 = 1141m/s.

    Vf = Vo(v) + gt,
    Solve for t and get:
    t = (Vf - Vo(v)) / g,
    t = (0 - = 1141) / 9.8 = 116.4s = t(up) = t(down).

    T = 2 * 116.4 = 232.8s = 3.9 min. =
    total flight time.

    Hor Range = Vo(h) * T,
    Hor Range = 1020m/s * 232.8s = 237,456m
    = 237.5km.

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