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A projectile is fired from the origin (at y = 0 m) as shown in the diagram. The initial velocity components are V0x = 310 m/s and V0y = 26 m/s. The projectile reaches maximum height at point P, then it falls and strikes the ground at point Q, which is 20 m below the launch point. For how long is the projectile in the air?

  • physics -

    Yo = 26 m/s.=Ver. component of initial velocity.

    Y = Yo + gt.
    Tr = (Y-Yo)/g = (0-26/-9.8 = 2.65 s. =
    Rise time.

    Tf1 = Tr = 2.65 s. = Fall time to launching point.

    h = Yo*t + 0.5g*t^2 = 20 m.
    26t + 4.9t^2 = 20
    4.9t^2 + 26t - 20 = 0
    Use Quad. Formula:
    Tf2 = 0.68 s. To fall 20 m.

    T = Tr + Tf1 + Tf2=2.65 + 2.65 + 0.68 =
    5.98 s. = Time in air.

  • physics -

    4.5s

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