Question 190294
<font face="Garamond" size="+2">


The height function is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  h(t) = -4.9t^2 + 245t + h_o]


Where *[tex \LARGE h_o] is the initial height of the projectile.  You could argue that if it were a 6' tall man firing the flare gun, then the initial height is about 2 meters, but I suspect that this little detail was ignored for this problem, so let's just set *[tex \LARGE h_o = 0].  That gives us a height function that looks like this:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  h(t) = -4.9t^2 + 245t ]


What this means is that you can calculate the instantaneous height for any value of <i>t</i>.  But that doesn't help directly because we are given the height and we need to discover the time.  Hence, the question becomes: "What is the value of <i>t</i> when the height is 1960 meters", or:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  -4.9t^2 + 245t = 1960]


Put the quadratic into standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  -4.9t^2 + 245t - 1960 = 0]


Solve with the quadratic equation:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \frac{-245 \pm sqrt{245^2 - 4(-4.9)(-1960)}}{2(-4.9)} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \frac{-245 \pm sqrt{21609}}{-9.8} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \frac{-245 \pm 147}{-9.8} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t_1 = 10]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t_2 = 40]


So the flare fired at time 0 gets to the balloonist at time 10 seconds and then passes him on the way back down at 40 seconds, i.e. 30 seconds later.


<i><b>Extra credit:</b></i>  How many seconds after the flare passes the balloonist does it hit the ground?


<i><b>Super-double-plus extra credit:</b></i>  When does the flare get as high as it will ever get, and how high is that?


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