Question 431079
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No.


The height of a projectile in meters, *[tex \Large t] seconds after being launched from an initial height of *[tex \Large h_o] at an initial vertical velocity of *[tex \Large v_o] is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -4.9t^2\ +\ v_ot\ +\ h_o]


presuming, of course, that all of this activity is occuring in the vicinity of the planet Earth.  I can't imagine where you came up with a -1.4 as a lead coefficient (too big for the moon and too small for Mercury or Mars...maybe one of Jupiter's moons) and I can't imagine how you can have distance as a function of time with a constant acceleration without having a squared variable.


Your problem has an initial velocity of 25 meters/second and an initial height of 12 meters, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -4.9t^2\ +\ 25t\ +\ 12]


You are concerned with finding the positive value of *[tex \Large t] when the projectile returns to the ground which is typically considered to be 0 height.


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ 0]


which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -4.9t^2\ +\ 25t\ +\ 12\ =\ 0]


Just use the quadratic formula to solve the quadratic for the positive root.


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