I am trying to understand projectile motion, but I have gotten stuck at the point at which they state in the limit $t >> v_t/g$ the equation is reduced, and when $t << v_t/g$ we also remove some of the equation. $>>$ means much greater than as far as I know and the other way means much less than, how do we quantify this?
I understand how the equations 184 and 187 were reached. But I do not understand why we then use 185, 186 and 188, 189?
Here is my source http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node29.html#e5.23u.
EDIT:
Also, 184 and 187 are functions that will return the $x$ and $z$ coordinates respectively given a time. I can calculate the time at which $z = 0$, i.e. when it hits the ground as that is equal to $(2v_0sin\theta)/g$. As such, the only unknown left is $v_t$. They say $v_t=m (g/c)$. Therefore, the unknown is $c$. What is $c$?
In the limit $t << v_t/g$, you may use the approximation $e^{-x}\approx 1-x$ to get
$$1- e^{-gt/v_t}\approx \frac{gt}{v_t}$$
and hence $x= v_0\cos\theta\> t $. (Eq. 185)
In the limit $t >> v_t/g$, use the approximate $e^{-gt/v_t}\rightarrow 0$ to get
$$1- e^{-gt/v_t}\approx 1$$
and hence $x = \frac{v_0v_t\cos\theta}{g}$. (Eq. 186)