Question 380857
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The function for the height of a projectile as a function of time, *[tex \Large t], given an initial velocity of *[tex \Large v_o] and an initial height of *[tex \Large h_o] is:


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


Substitute the known values to get the specific function for this problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -16t^2\ +\ 45t\ +\ 664].


The height of the ground is zero, so if we want to know when the projectile hits the ground, *[tex \Large h(t)\ =\ 0].


Therefore solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16t^2\ +\ 45t\ +\ 664\ =\ 0]


for *[tex \Large t] and discard the negative root.


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