Question 1158564
<font face="Times New Roman" size="+2">


The height as function of elapsed time for an object's vertical motion near the earth's surface using the fps measurement system is:


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


Where *[tex \Large h(t)] represents the height at time *[tex \Large t], *[tex \Large -16] is one-half of the gravitational acceleration constant near the earth, *[tex \Large v_o] is the initial velocity of the object, and *[tex \Large h_o] is the initial height of the object.


For both experiments, the initial height, *[tex \Large h_o\ =\ 300\text{ feet}], and the gravitational acceleration constant is, as the name implies, a constant.


For one of these hooligans showing egregious disrespect for their algebra textbooks, the initial velocity is *[tex \Large v_o\ =\ -10\text{ f/s}]. This value is negative because "...<b>towards the ground</b>...". For the other, the initial velocity is zero.


When the books hit the ground, the height is zero. Hence, you need to solve the following two quadratic equations to determine the values of *[tex \Large t_T], the elapsed time for the book thrown and *[tex \Large t_D], the elapsed time for the book dropped so that you can make the final calculation of *[tex \Large t_D\ -\ t_T], which answers the question posed.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16\(t_T\)^2\ -\ 10t_T\ +\ 300\ = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -16\(t_D\)^2\ +\ 300\ = 0]


The rest is just arithmetic.


<b><i>Extra Credit</i></b>


Which book hits the ground first if one just drops his and the other throws his UPWARDS at 10 feet per second?
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>