The height y (in feet) of a ball thrown by a child is y=−1/12x^2+6x+3 where x is the horizontal distance in feet from

Question

The height y (in feet) of a ball thrown by a child is

y=−1/12x^2+6x+3

where x is the horizontal distance in feet from the point at which the ball is thrown.

(a) How high is the ball when it leaves the child’s hand? (Hint: Find y when x=0)

Your answer is y=_______

(b) What is the maximum height of the ball? _______

(c) How far from the child does the ball strike the ground? ______

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Gabriella 2 weeks 2021-09-11T01:03:30+00:00 1 Answer 0

Answers ( )

    0
    2021-09-11T01:05:06+00:00

    Answer:

    Step-by-step explanation:

    a)

    y=−1/12x^2+6x+3

    y=−1/12(0)^2+6(0)+3

    y = 3

    b)

    y=−1/12x^2+6x+3

    y = -1/12 (x^2-72x) + 3

    y = =-1/12 (x^2-72x+1296-1296) +3

    y = -1/12(x^2 -72x +1296) + 108 + 3

    y = -1/12 (x – 36)^2 +111

    maximum height of the ball is 111 feets

    c)

    y = -1/12 (x – 36)^2 +111

    0 = -1/12 (x – 36)^2 +111

    -111 = -1/12 (x – 36)^2

    1332 = (x – 36)^2

    36.497 = x – 36

    x = 72.497

    How far from the child does the ball strike the ground = 72.497 feets

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