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26 April, 02:01

The height, in feet, after x seconds of an object launched straight up can be found by the function h (x) = - 16x2+v0x+h0, where v0 is the initial velocity of the object and h0 is the initial height.

A ball is kicked straight up into the air from a height of 12 ft with an initial velocity of 44 ft/s. After how many seconds does the ball hit the ground?

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  1. 26 April, 02:47
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    The function is:

    h (x) = - 16 x² + vo x + h o

    v o = 44 ft / s (initial velocity)

    h o = 12 ft (initial height)

    - 16 x² + 44 x + 12 = 0 / : 4 (we will divide both sides by 4)

    - 4 x² + 11 x + 3 = 0

    - 4 x² + 12 x - x + 3 = 0

    - 4 x (x - 3) - (x - 3) = 0

    (x - 3) ( - 4 x - 1) = 0

    x - 3 = 0, x = 3 s

    or:

    - 4 x - 1 = 0, - 4 x = 1, x = - 1/4 (false).

    Answer:

    3 seconds.
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