Question 1121393
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Presuming that the person who threw the ball was standing in a hole of such depth that when the ball was released it was exactly at ground level, then your function makes sense.


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


1.  Evaluate *[tex \LARGE h(1)].


2.  Set *[tex \LARGE h(t)\ =\ 0], that is *[tex \LARGE t(96\ -\ 16t)\ =\ 0] and then solve for *[tex \LARGE t].  Discard the zero root because 0 is the time the ball was thrown.


3.  Set *[tex \LARGE h(t)\ =\ 40], that is *[tex \LARGE t(96\ -\ 16t)\ =\ 40] and then solve for *[tex \LARGE t].  There will be two roots, one representing the time the ball was at 40 feet from the ground on the way up, and one representing the time it was at 40 feet from the ground on the way down.


Extra credit:  What is the highest point that the ball reaches? In other words, does it make sense that the ball was at 40 ft above the ground twice during its travels?  Did it ever get to 40 ft high at all?

Hint:  Put your function into *[tex \LARGE h(t)\ =\ at^2\ +\ bt\ +\ c] form, then calculate *[tex \LARGE -\frac{b}{2a}] and evaluate the function at that value.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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