Question 1158111
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The proper way to express this function in general is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\frac{1}{2}gt^2\ +\ v_ot\ +\ h_o]


Where *[tex \Large g\ =\ -32\,\text{ft/sec^2}], the vertical acceleration of an object near the earth's surface, *[tex \Large v_o] is the initial velocity, and *[tex \Large h_o] is the initial height.


In your case, *[tex \Large v_o\ =\ 96\,\text{ft/sec}] and *[tex \Large h_o\ =\ 0\,\text{ft}]


The projectile is on the ground whenever the value of the function is zero, so *[tex \Large h(t_o)\ =\ h(0)\ =\ 0] means the ball is on the ground at the time of launch, and *[tex \Large h(t_f)\ =\ 0] means that the ball returns to the ground at time *[tex \Large t_f].  


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -16t^2\ +\ 96t]


Has two zeros, one at time zero or *[tex \Large t_o\ =\ 0].  You can solve the quadratic equation


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16t^2\ +\ 96t\ =\ 0]


for the other.


The graph of the function is a parabola opening downward.  The maximum height is the maximum value of the height function at the vertex of the parabola.  The time in seconds for the ball to reach this height is the value of the independent variable at the vertex of the parabola.  In other words, the vertex of the parabola is at the point *[tex \Large \(t,\,h(t))] where the ordinate of the point is the maximum height, and the abscissa is the time in seconds it takes to get there.


Hint:  The vertex of a parabola of the form *[tex \Large f(x)\ =\ ax^2\ +\ bx\ +\ c] is found at the point *[tex \Large \(-\frac{b}{2a},f\(-\frac{b}{2a}\)\)]


That is all the information you need to solve this problem for yourself.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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