Question 1178877
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This is a quadratic function with a negative lead coefficient, therefore the graph of the function is a concave down parabola with a vertex at *[tex \Large \(t_{max},h(t_{max})\)].


For a general quadratic in *[tex \Large t] where *[tex \Large a\ <\ 0], namely *[tex \Large at^2\ +\ bt\ +\ c], the value of *[tex \Large t_{max}] is given by *[tex \Large -\frac{b}{2a}].  And *[tex \Large h(t_{max})\ =\ h\(-\frac{b}{2a}\)]


The ball hits the ground at the positive value of *[tex \Large t] such that *[tex \Large h(t)\ =\ 0].  Set the function equal to zero and solve for the positive root.


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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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