Dynamics

A level fight bomber flying at 300 ft/s releases a bomb at an elevation of 6400 ft. Approximately how long before the bomb strikes the earth?

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  1. Only the vertical component counts for the time for descent.
    Assuming no air resistance, we have from kinematics,
    S=ut+(1/2)at^2
    where
    S=distance = 6400 ft
    u=initial velocity (=0 ft/s vertical in this case)
    a=acceleration due to gravity, g=32.2 ft/s²

    Substitute values,
    6400=(1/2)(32.2)t²

    Solve for t.

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  2. T=20s

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  3. what is the formula

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