Question 234663
<font face="Garamond" size="+2">


The vertical distance as a function of time is given by:


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


where *[tex \Large g = 9.8] m/secē is the acceleration due to gravity, *[tex \Large v_o\ = 20] m/sec is the initial velocity, and *[tex \Large h_o] is the height of your hand above the ground when you let go of the rock.


First answer part b).  To answer part b), realize that the graph of this function is a parabola opening downward with vertex at


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(t_{max},\,h(t_{max})\right)]


and that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t_m\ =\ \frac{-v_o}{2\,\cdot\,\frac{g}{2}}\ =\ -\frac{v_o}{g}]


because for a general quadratic function, *[tex \Large f(x)\ =\ ax^2\ +\ bx\ +\ c], the *[tex \Large x]-coordinate of the vertex is *[tex \Large \frac{-b}{2a}]


Once you have calculated *[tex \Large t_{max}], which is the solution to part b), find the value of the height function for that time, that is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t_{max})\ =\ -\frac{1}{2}g(t_{max})^2\ +\ v_ot_{max}\ +\ h_o]


Making the appropriate substitution for *[tex \Large h_o] based on your personal physical stature and the mechanics of your throwing arm and you will have the solution to part a) 


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>