# The falling body

• Oct 15th 2007, 06:00 PM
chrsr345
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
• Oct 15th 2007, 06:33 PM
Jhevon
Quote:

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

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

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

can you continue?
• Oct 16th 2007, 09:48 AM
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...
• Oct 16th 2007, 09:59 AM
topsquark
Quote:

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...

$\displaystyle v = \frac{mg}{c} - \frac{mg}{c}e^{-ct/m}$

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

-Dan
• Oct 16th 2007, 10:38 AM
chrsr345
Quote:

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

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

-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...
• Oct 16th 2007, 01:50 PM
topsquark
Quote:

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

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

-Dan

Quote:

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 $\displaystyle a^{-b} = \frac{1}{a^b}$, so our expressions are the same. Unless you meant to write
$\displaystyle v(t) = -\frac{1}{c} \cdot \frac{mg}{e^{ct/m} - mg}$
(which would be incorrect.)

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

-Dan
• Oct 16th 2007, 03:58 PM
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...:confused:
• Oct 16th 2007, 04:05 PM
topsquark
Quote:

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...:confused:

Does this help?
$\displaystyle v = \frac{mg}{c} - \frac{mg}{c}e^{-ct/m}$

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

-Dan