Question 608152
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Well, not <i><b>exactly</b></i>.


Your height function ignores the initial height of the tip of the arrow at the time of release.  This value is other than zero unless the archer is standing in a hole of sufficient depth such that the tip of the arrow is exactly at ground level at the time it is released.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -16t^2\ +\ 160t\ +\ h_o]


Where *[tex \LARGE h_o] is the initial height.  We'll just accept the "archer in a hole" scenario for the time being and assume that *[tex \LARGE h_o\ =\ 0].


The graph of your function is a parabola.  Because of the negative lead coefficient, the parabola opens downward, meaning that the vertex is a maximum value of the function.  The *[tex \LARGE x] coordinate of the vertex of a parabola described by the function *[tex \LARGE \rho(x)\ =\ ax^2\ +\ bx\ +\ c] such as this one is given by:


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


Calculate *[tex \LARGE x_v]


Once you have a value for the *[tex \LARGE x] coordinate of the vertex, calculate


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(x_v)]


to get the value of the function at the vertex, i.e. the maximum value of the height function.  If you decide later that *[tex \LARGE h_o] is actually something other than zero, you can just add it on at the end.


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