The height of a ball thrown into the air after t seconds have elapsed is h = −16t2 + 40t + 6 feet. What is the first time, t, when the ball

Question

The height of a ball thrown into the air after t seconds have elapsed is h = −16t2 + 40t + 6 feet. What is the first time, t, when the ball will reach a height of 20 feet? Round your answer to two decimal places.

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Katherine 3 days 2021-11-25T13:06:38+00:00 1 Answer 0 views 0

Answers ( )

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    2021-11-25T13:07:55+00:00

    The first time when the ball will reach a height of 20 feet is 0.42 seconds

    Solution:

    Given that,

    The height of a ball thrown into the air after t seconds have elapsed is:

    h = -16t^2 + 40t + 6

    What is the first time, t, when the ball will reach a height of 20 feet?

    Substitute h = 20

    20 = -16t^2 + 40t + 6\\\\-16t^2 + 40t + 6 -20 = 0\\\\-16t^2 + 40t -14 = 0\\\\16t^2 -40t + 14 = 0\\\\8t^2 -20t + 7=0

    Solve by quadractic formula

    \mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}

    x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

    \mathrm{For\:}\quad a=8,\:b=-20,\:c=7

    t = \frac{-\left(-20\right)\pm \sqrt{\left(-20\right)^2-4\cdot \:8\cdot \:7}}{2\cdot \:8}\\\\t = \frac{20 \pm \sqrt{176}}{16}\\\\t = \frac{20 \pm 4\sqrt{11}}{16}\\\\t = \frac{ 5 \pm \sqrt{11}}{4}\\\\We\ have\ two\ solutions\\\\ t=2.07915, \:t=0.42084

    Rounding off we get,

    t = 2.08 , t = 0.42

    Thus the first time when the ball will reach a height of 20 feet is 0.42 seconds

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