Question 235484
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If you graph the function you will see that it is a parabola opening downward -- which is to be expected because it is a quadratic function with a negative lead coefficient.


The maximum height will be reached at the *[tex \Large t] coordinate of the vertex of the parabola.  For the general quadratic function, *[tex \Large f(x)\ =\ ax^2\ +\ bx\ + c], the *[tex \Large x]-coordinate of the vertex is found by:


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


For your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t_{max} = \frac{-200}{2(-16)}]


You can do your own arithmetic.


The maximum height is the value of the function at time *[tex \Large t_{max}], that is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s(t_{max})\ =\ -16(t_{max})^2\ +\ 200(t_{max})\ +\ 4]


Again, the arithmetic is yours to perform.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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