Physics

A brick is thrown upward from the top of a building at an angle of 35° to the horizontal and with an initial speed of 12 m/s. If the brick is in flight for 3.5 s, how tall is the building?

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  1. Vi = 12 sin 35 = 6.88 m/s

    how far above building does it go?
    V top = 0 = 6.88 - 9.8 t
    t = .7 seconds up
    ke = 0 at top = (1/2) v^2 - gh
    .5(6.88)^2 = 9.8 h
    h = 2.42 meters above building

    now, new problem. drop brick from 2.42 meters above building
    time falling = 3.5 - .7 = 2.8 seconds
    h = 4.9 (2.8)^2
    = 38.4 from top
    so building is 38.4 - 2.4 = 36 meters high

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