Question 327281
<font face="Garamond" size="+2">


You throw an egg into the air.  When the egg leaves your hand at 48 feet per second initial velocity, the egg is 6 feet above the ground.  When the egg splatters on the ground, you expect it to make a mess that is 8 inches in diameter.  At the moment you throw the egg, an ant moving at 0.2 feet per second along the ground passes the exact spot that represents the center of the circular deposit of egg contents.  Will the ant be inside or outside of the radius of the splatter at the time the egg impacts the ground?


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -16t^2\ +v_ot\ +\ h_o]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -16t^2\ +48t\ +\ 6]


Height of the ground is 0 feet, so the time to reach 0 feet is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16t^2\ +48t\ +\ 6\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}]


where *[tex \LARGE a\ =\ -16], *[tex \LARGE b\ =\ 48], and *[tex \LARGE c\ =\ 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{-48\ \pm\ \sqrt{48^2\ -\ 4(-16)(6)}}{2(-16)}]


This simplifies to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{6\ \pm\ \sqrt{42}}{8}]


Discard the negative root (who can go back in time?)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{6\ +\ \sqrt{42}}{8}\ \approx\ 3.12]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3.12 \times\ 0.2\ =\ 0.624] feet is the distance the ant traveled.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8 in\ =\ \frac{8}{12}\ =\ 0.667] feet.  Got him!


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