Question 441499
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You have a projectile launched vertically at 80 feet per second initial velocity from a position 4 feet above ground level (height zero).


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -16t^2\ +\ 80t\ +\ 4]


gives the height at *[tex \Large t] seconds if the instant of launch is *[tex \Large t\ =\ 0]


You want to know the value of *[tex \Large t] when *[tex \Large h(t)\ =\ 100]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16t^2\ +\ 80t\ +\ 4\ =\ 100]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16t^2\ +\ 80t\ -\ 96\ =\ 0]


Just solve the quadratic for the two values of *[tex \Large t].  Hint:  Divide through by -16 and then factor the resulting trinomial.


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