physics

A brick is thrown upward from the top of a building at an angle of 25° to the horizontal and with an initial speed of 16 m/s. If the brick is in flight for 3.1 s, how tall is the building?

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asked by sid
  1. The vertical component of the speed is initially
    Voy = 16 sin25 = 6.762 m/s.

    Height above ground is

    y = H + Voy*t - (g/2)t^2

    y = H +6.762t -4.9 t^2

    You know that y = 0 when t = 3.1 s.

    Solve for H, the building height.

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    posted by drwls

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