Question 439691
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Your function makes no sense at all in the physical world.


Solving it for t when h = 0 yields 3/5 seconds and 6 seconds.  How can something launched from the top of a hill reach 0 height twice?


Furthermore, if you multiply the two binomials you get 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -5t^2\ +\ 33t\ -\ 18]


Though you failed to mention what units the height is measured in.


The function for the height of a projectile launched near the surface of any celestial body is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -\frac{1}{2}g_it^2\ +\ v_ot\ +\ h_o]


Where *[tex \Large g_i] is the acceleration due to gravity of the selected celestial body, *[tex \Large 32\ ft/sec^2] or *[tex \Large 9.8\ m/sec^2], in the case of Earth, *[tex \Large v_o] is the initial velocity, and *[tex \Large h_o] is the initial height.  


If you assume that you are using the mks system, then your gravitational acceleration, assuming you took very inappropriate roundoff liberties would allow for being on Earth.  I don't know where you are if you are using fps.


But the problem is the anomaly caused by the fact that you are claiming to have launched from the top of a hill, but your initial height is a negative number.  Granted, I am only 62 years old, so my experience is somewhat limited, but every hill in my experience has had a positive value for the height.


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