Question 620999
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The height of a projectile at time *[tex \LARGE t] is given by the function


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


Where *[tex \LARGE g] is the acceleration due to gravity near the surface of whatever planet you are on.  *[tex \LARGE 32\ \text{\frac{ft}{sec^2}}] if you are still on Earth.  *[tex \LARGE v_o] is the initial upward velocity in feet per second, and *[tex \LARGE h_o] is the initial height in feet.


The ground is height zero.  Set the function equal to zero and solve for *[tex \LARGE t].  Use the quadratic formula.  Discard the negative root.


Since acceleration due to gravity is a negative quantity with respect to the initial direction of the rocket, the graph of the function is a parabola opening downward.  Hence the vertex of the parabola is at the point *[tex \LARGE \left(t_v,h(t_v)\right)], where *[tex \LARGE t_v] represents the time that the vertex is obtained and *[tex \LARGE h(t_v)] is the maximum height obtained.  Recall that the vertex of the parabola *[tex \LARGE \rho(x)\ =\ ax^2\ +\ bx\ +\ c] has an *[tex \LARGE x]-coordinate of *[tex \LARGE \frac{-b}{2a}]


Set the function equal to 50 and solve the quadratic for *[tex \LARGE t].  Use the quadratic formula.


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