Question 334054
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -16t^2\ +\ v_ot\ +\ h_o]


where *[tex \Large v_0\ =\ 32] and *[tex \Large h_0\ =\ 8]


a) calculate *[tex \Large h(1)\ =\ -16(1)^2\ +\ 32(1)\ +\ 8]


b) set *[tex \Large h(t)\ =\ 0] and solve the quadratic for *[tex \Large t]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16t^2\ +\ 32t\ +\ 8\ =\ 0]


c) calculate the *[tex \Large t]-coordinate of the vertex of your parabola using:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-b}{2a}\ =\ \frac{-32}{-32}\ =\ 1]


Then calculate the value of the function at that time value, i.e. *[tex \Large h(1)\ =\ -16(1)^2\ +\ 32(1)\ +\ 8] (Note: same answer as part a)


Public Service Notice:  No animals, in particular poodles, were harmed during the calculation of the solution to this problem.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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