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23 January, 01:33

A ball is thrown upward and outward from a height of 77 feet. the height of the ball, f (x), in feet, can be modeled by f left parenthesis x right parenthesis equals negative 0.2 x squared plus 1.4 x plus 7f (x) = - 0.2x2+1.4x+7 where x is the ball's horizontal distance, in feet, from where it was thrown. use this model to solve parts (a) through (c).

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  1. 23 January, 01:39
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    From the given function modeling the height of the ball:

    f (x) = - 0.2x^2+1.4x+7

    A] The maximum height of the ball will be given by:

    At max height f' (x) = 0

    from f (x),

    f' (x) = - 0.4x+1.4

    solving for x we get:

    -0.4x=-1.4

    x=3.5ft

    thus the maximum height would be:

    f (3.5) = - 0.2 (3.5) ^2+1.4 (3.5) + 7

    f (3.5) = 9.45 ft

    b]

    How far from where the ball was thrown did this occur:

    from (a), we see that at maximum height f' (x) = 0

    f' (x) = - 0.4x+1.4

    solving for x we get:

    -0.4x=-1.4

    x=3.5ft

    This implies that it occurred 3.5 ft from where the ball was thrown.

    c] How far does the ball travel horizontally?

    f (x) = - 0.2x^2+1.4x+7

    evaluationg the expression when f (x) = 0 we get:

    0=-0.2x^2+1.4x+7

    Using quadratic equation formula:

    x=-3.37386 or x=10.3739

    We leave out the negative and take the positive answer. Hence the answer 10.3739 ft horizontally.
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