Question 777517
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a)  Solve *[tex \LARGE -16t^2\ +\ 40t\ =\ 24].  You will get two zeros for this equation, one will represent the time the ball reaches 24 ft on the way up, and the other will represent the time the ball reaches 24 ft on the way back down.


b)  Solve *[tex \LARGE -16t^2\ +\ 40t\ =\ 48].  This equation has a conjugate pair of complex zeros.  What do you suppose that means?  Wait to answer this one until after you have answered c) and d).


Do d) before you do c)


d)  The *[tex \LARGE x]-coordinate of the vertex of *[tex \LARGE f(x)\ =\ ax^2\ +\ bx\ +\ c] is given by *[tex \LARGE \frac{-b}{2a}].  So calculate *[tex \LARGE \frac{-40}{2(-16)}].  This is the time of the highest point.


c)  Use the result of part d), *[tex \LARGE \left(t_{max}\right)], to calculate *[tex \LARGE h] at time *[tex \LARGE t_{max}].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h_{max}\ =\ -16\left(t_{max}\right)^2\ +\ 40t_{max}]


NOW answer the question: "When does the ball reach a height of 48 ft?"


e)  Solve *[tex \LARGE -16t^2\ +\ 40t\ =\ 0] for the non-zero root.  (time zero is when you threw the ball in the first place.)


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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