Question 319109
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The more general formula is:


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


Where *[tex \Large h] is the height as a function of time, *[tex \Large t], *[tex \Large v_o] is the initial velocity, and *[tex \Large h_o] is the initial height.


For your problem the rocket was launched from the ground so *[tex \Large h_o\ =\ 0], and *[tex \Large v_o\ =\ 560] fps.


We need to know the value of *[tex \Large t] that makes *[tex \Large h(t)\ =\ 3,136] feet.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16t^2\ +\ 560t\ =\ 3136]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16t^2\ +\ 560t\ -\ 3136\ =\ 0]


Divide through by *[tex \Large -16]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t^2\ -\ 35t\ +\ 196\ =\ 0]


This quadratic factors quite neatly, yielding two positive roots.  I'll leave it as an exercise for the student to figure out the significance of two answers, both apparently valid.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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