Question 971165
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Your height function is incorrect.  The function you provided is the height ABOVE the roof of the house. The correct function is:


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


where *[tex \Large v_o] is the initial velocity and *[tex \Large h_o] is the initial height of the projectile when it was fired.  Since each floor of the hous measures 12 feet, one must presume that this value is equal to 24 feet.  Your function is therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ H(t)\ =\ -16t^2\ +\ 44t\ +\ 24]


<u>Algebra method</u>


The height function is a quadratic polynomial that has a graph that is a parabola.  Since the lead coefficient is negative, the parabola opens downward and the *[tex \Large H(t)] coordinate of the vertex represents the maximum height of the projectile and the *[tex \Large t] coordinate of the vertex represents the time at which the projectile reaches that maximum height.


The coordinates of the vertex of 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ ax^2\ +\ bx\ +\ c]


are given by


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_v = f(x_v) = f\left(\frac{-b}{2a}\right)]


So for your situation


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t_{max} = \frac{-44}{2(-16)}]


You can do your own arithmetic to find *[tex \Large t_{max}]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ H(t_{max})\ =\ -16(t_{max})^2\ +\ 44t_{max}\ +\ 24]


Just substitute and calculate.


<u>Calculus Method</u>


The local maximum is the point where the first derivative is equal to zero, so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ H(t)\ =\ -16t^2\ +\ 44t\ +\ 24]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dH}{dt}\ =\ -32t\ +\ 44]


Set *[tex \Large -32t\ +\ 44\ =\ 0]


And solve for *[tex \Large t]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ H(t)\ =\ -16t^2\ +\ 44t\ +\ 24]


Just substitute and calculate.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \