1. The falling body

In the last of my relatively hard problems to figure out, this one applies to a falling body.

ln(mg / mg-cv) = (c/m) t

I need to solve for v(t) and what happens as t-->∞

Thanks, and yes im also new to ln so im a little embarassed if this is really easy. Thanks

Chris

2. Originally Posted by chrsr345
In the last of my relatively hard problems to figure out, this one applies to a falling body.

ln(mg / mg-cv) = (c/m) t

I need to solve for v(t) and what happens as t-->∞

Thanks, and yes im also new to ln so im a little embarassed if this is really easy. Thanks

Chris
$\ln \left( \frac {mg}{mg - cv} \right) = \frac cmt$

$\Rightarrow \frac {mg}{mg - cv} = e^{\frac cmt}$

can you continue?

3. Ok thanks again jhevon. What i came up with looks...messy.

V(t) = -1/C (mg/ e^((c/m)t) - mg)

And if this is right, i have no idea what this looks like as it approaches infinity...

4. Originally Posted by chrsr345
Ok thanks again jhevon. What i came up with looks...messy.

V(t) = -1/C (mg/ e^((c/m)t) - mg)

And if this is right, i have no idea what this looks like as it approaches infinity...
$v = \frac{mg}{c} - \frac{mg}{c}e^{-ct/m}$

So as t approaches infinity, the argument of the exponential becomes very large and negative.
$\lim_{x \to \infty}e^{-x} =$?

-Dan

5. Originally Posted by topsquark
$v = \frac{mg}{c} - \frac{mg}{c}e^{-ct/m}$

So as t approaches infinity, the argument of the exponential becomes very large and negative.
$\lim_{x \to \infty}e^{-x} =$?

-Dan
Dan, is that equation what you got or is it written version of my answer? because mine is a little bit different, i have the e^((c/m)t) in the denominator. Maybe i forgot a set of parenthesis...

V(t) = -1/C ((mg/ e^((c/m)t)) - mg)..

OR = {-mg / Ce^[(c/m)t]} + {mg / C}

I could take a pic of my work if that makes things easier...

6. Originally Posted by topsquark
$v = \frac{mg}{c} - \frac{mg}{c}e^{-ct/m}$

So as t approaches infinity, the argument of the exponential becomes very large and negative.
$\lim_{x \to \infty}e^{-x} =$?

-Dan
Originally Posted by chrsr345
Dan, is that equation what you got or is it written version of my answer? because mine is a little bit different, i have the e^((c/m)t) in the denominator. Maybe i forgot a set of parenthesis...

V(t) = -1/C ((mg/ e^((c/m)t)) - mg)..

OR = {-mg / Ce^[(c/m)t]} + {mg / C}

I could take a pic of my work if that makes things easier...
Recall that $a^{-b} = \frac{1}{a^b}$, so our expressions are the same. Unless you meant to write
$v(t) = -\frac{1}{c} \cdot \frac{mg}{e^{ct/m} - mg}$
(which would be incorrect.)

Either way the question you need to answer is the same. I'm asking you for the value of
$\lim_{x \to \infty} e^{-x} = \lim_{x \to \infty} \frac{1}{e^x}$
(This last form might be easier for you to consider.)

-Dan

7. Ah yes, that is the same answer i got, thank you!
And as for the limit...i would guess that as t approaches infinity, that thew function would get infinitely small as its in the denominator....Although with the final equation, its hard for me to picture this in my head...

8. Originally Posted by chrsr345
Ah yes, that is the same answer i got, thank you!
And as for the limit...i would guess that as t approaches infinity, that thew function would get infinitely small as its in the denominator....Although with the final equation, its hard for me to picture this in my head...
Does this help?
$v = \frac{mg}{c} - \frac{mg}{c}e^{-ct/m}$

So
$\lim_{t \to \infty} \left ( \frac{mg}{c} - \frac{mg}{c}e^{-ct/m} \right ) = \frac{mg}{c} - \frac{mg}{c} \cdot 0 = \frac{mg}{c}$

-Dan