GRAVITATION The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocit

Question

GRAVITATION The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation h(t) = –16t2 + 60t. At what times will the object be at a height of 56 feet?
A. 14 s, 6 s
B. 1.75 s, 2 s
C. 8 s, 1 s
D. −1.75 s, −2 s

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Mackenzie 2 weeks 2021-11-21T12:23:23+00:00 1 Answer 0 views 0

Answers ( )

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    2021-11-21T12:25:02+00:00

    Option B

    At 2 second and 1.75 second, the object be at a height of 56 feet

    Solution:

    Given that,

    The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation:

    h(t) = -16t^2 + 60t

    At what times will the object be at a height of 56 feet

    Substitute h = 56

    56 = -16t^2 + 60t\\\\16t^2 - 60t + 56 = 0\\\\Divide\ the\ equation\ by\ 4\\\\4t^2 - 15t + 14=0

    Solve the above equation by quadratic 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=4,\:b=-15,\:c=14\\\\x =\frac{-\left(-15\right)\pm \sqrt{\left(-15\right)^2-4\cdot \:4\cdot \:14}}{2\cdot \:4}\\\\Simplify\\\\x = \frac{15 \pm \sqrt{1}}{8}\\\\x = \frac{15 \pm 1}{8}\\\\We\ have\ two\ solutions\\\\x = \frac{15+1}{8} \text{ and } x = \frac{15-1}{8}\\\\x = 2 \text{ and } 1.75

    Thus, at 2 second and 1.75 second, the object be at a height of 56 feet

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