Question 596989
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Your equation is only accurate to the extent that the ball is released when it is 5 feet above the ground and it is thrown upward with a vertical initial velocity component of 11 feet per second.  In general, a projectile propelled upward at an initial velocity of *[tex \LARGE v_o] distance units per time unit from an initial height of *[tex \LARGE h_o] distance units above the surface of a celestial body with a near-surface gravitational acceleration of *[tex \LARGE g_i] distance units per time unit per time unit is:


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


Other than that, what is it you wanted to know?


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