Why projectile motion equations can be reduced in certain intervals

Question

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 $c$?

score 0 · Accepted Answer · answered Oct 26 '19 at 14:23

0

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)

answered Oct 26 '19 at 14:23

Quanto

97,352

Thanks. Do you know how I can find $c$ such that I can actually calculate some coordinates given $\theta$, $v_0$ and $m$? – F J Oct 26 '19 at 14:27
Is $c$ momentum? I.e. $c = mv_0$ – F J Oct 26 '19 at 14:39
$c$ is the drag coefficient, which is usually given to determine the terminal velocity $v_t = mg/c$. Conversely, $c$ can be obtained if $v_t$ is known. – Quanto Oct 26 '19 at 15:43
I need $c$ to calculate $v_t$, but I need $v_t$ to calculate $c$... How do I actually calculate any of the coordinates then? – F J Oct 26 '19 at 15:45
$v_t$ is in $m/s$, and so I want the right side of the equation to be in $m/s$. Since $v_t = m g/c$, we have $kgm/s^2$ on the top and as such $c$ must be in $kgm/s$ which is momentum. So is $c = mv_0$? – F J Oct 26 '19 at 15:47
One of them should be a given in the problem. – Quanto Oct 26 '19 at 15:54
I'm not trying to solve someone's problem, I'm trying to work it out for any value myself. So, how do I determine $c$? – F J Oct 26 '19 at 15:55
c is not equal to m$v_0$ – Quanto Oct 26 '19 at 15:55
Then what is it? – F J Oct 26 '19 at 15:57
you need to know v_t to obtain c. In case the time of flight t is known, you could calculate v_t using 187 by setting z=0, as you described in the Edit. – Quanto Oct 26 '19 at 16:21
So plug in $t = t_f$ and $z = 0$ then solve for $v_t$? – F J Oct 26 '19 at 16:28
Right, except it is a non-linear equation to solve. – Quanto Oct 26 '19 at 16:41
Yeah I'm having a lot of trouble doing it. How can I solve it for $v_t$? – F J Oct 26 '19 at 16:42
There may not be an analytical solution; you may have to solve it numerically – Quanto Oct 26 '19 at 16:52
I don't know what that means – F J Oct 26 '19 at 16:52
Would it not make more sense to solve for $v_t$ in 180? – F J Oct 26 '19 at 17:14

Why projectile motion equations can be reduced in certain intervals

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