M2215b.14 Find the limiting velocity

[karush]

$\text{14 The velocity of a body of mass $m$ falling from rest of gravity is given by the equation}\\$
\begin{align*}\displaystyle
V&=\sqrt{\frac{mg}{k}}\tanh\left[\frac{gk}{m} \right]
\end{align*}
$\text{$k$ is a constant that depends on the bodys aerodynamic properties \\ and the density of the air}\\$
$\text{$g$ is the gravitational constant}\\$
$\text{$t$ is the number of seconds into the fall.}\\$
$\text{Find the limiting velocity}$\begin{align*}\displaystyle
&\lim_{t \to \infty}v
\end{align*}The choices are
A) There is no limiting g speed
B) 0.01 ft/sec
C) 177.95 ft/sec
D) 56.27 ft/secIm not real sure how to set this up except to set $k$ to zero or just plain absent
then we are dealing with the constant of gravitation pull

 

[MarkFL]

$\text{14 The velocity of a body of mass $m$ falling from rest of gravity is given by the equation}\\$
\begin{align*}\displaystyle
V&=\sqrt{\frac{mg}{k}}\tanh\left[\frac{gk}{m} \right]
\end{align*}

You've given a constant velocity here...

 

[karush]

I really don't see how you can derive any ft/sec from this

 

[Greg]

$\text{The velocity of a body of mass $m$ falling from rest of gravity is given by the equation}\\$
\begin{align*}\displaystyle
V&=\sqrt{\frac{mg}{k}}\tanh\left[\frac{gk}{m} \right]
\end{align*}
$\text{$k$ is a constant that depends on the bodys aerodynamic properties \\ and the density of the air}\\$
$\text{$g$ is the gravitational constant}\\$
$\text{$t$ is the number of seconds into the fall.}\\$

Where is $t$?

 

[karush]

Where is $t$?

https://www.physicsforums.com/attachments/7627
ok appartently i left it off
i have had no physics so this is
very new
don't know where the numbers come from

 

[MarkFL]

Okay, now the given velocity function is dimensionally consistent, given that $k$ has units of mass per length (can you demonstrate this?).

In order to find the limiting or terminal velocity $v_T$, you need to find:

\(\displaystyle v_T=\lim_{t\to\infty}v(t)\)

What is:

\(\displaystyle \lim_{u\to\infty}\tanh(u)\)

Hint: Rewrite the hyperbolic tangent function in terms of the exponential definition...

 

[karush]

Okay, now the given velocity function is dimensionally consistent, given that $k$ has units of mass per length (can you demonstrate this?).

In order to find the limiting or terminal velocity $v_T$, you need to find:

\(\displaystyle v_T=\lim_{t\to\infty}v(t)\)

What is:

\(\displaystyle \lim_{u\to\infty}\tanh(u)\)

Hint: Rewrite the hyperbolic tangent function in terms of the exponential definition...

this? via W|A

$\displaystyle\frac{e^{\sqrt{x}}}
{e^{-\sqrt{x}}+e^{\sqrt{x}}}
-\frac{e^{-\sqrt{x}}}
{e^{-\sqrt{x}}+e^{\sqrt{x}}} $

 

[MarkFL]

this? via W|A

$\displaystyle\frac{e^{\sqrt{x}}}
{e^{-\sqrt{x}}+e^{\sqrt{x}}}
-\frac{e^{-\sqrt{x}}}
{e^{-\sqrt{x}}+e^{\sqrt{x}}} $

No, I was thinking more along the lines of:

\(\displaystyle \tanh(u)\equiv\frac{e^{u}-e^{-u}}{e^{u}+e^{-u}}\)

What happens if you multiply that by:

\(\displaystyle 1=\frac{e^{-u}}{e^{-u}}\,?\)

 

[karush]

No, I was thinking more along the lines of:

\(\displaystyle \tanh(u)\equiv\frac{e^{u}-e^{-u}}{e^{u}+e^{-u}}\)

What happens if you multiply that by:

\(\displaystyle 1=\frac{e^{-u}}{e^{-u}}\,?\)

$\displaystyle \tanh(u)\equiv\frac{e^{u}-e^{-u}}{e^{u}+e^{-u}}
\cdot\frac{e^{-u}}{e^{-u}}=\frac{1-e^{-2u}}{1+e^{-2u}}$

 

[MarkFL]

$\displaystyle \tanh(u)\equiv\frac{e^{u}-e^{-u}}{e^{u}+e^{-u}}
\cdot\frac{e^{-u}}{e^{-u}}=\frac{1-e^{-2u}}{1+e^{-2u}}$

Yes, so what's the limit:

\(\displaystyle \lim_{u\to\infty}\tanh(u)\,?\)

 

[karush]

Yes, so what's the limit:

\(\displaystyle \lim_{u\to\infty}\tanh(u)\,?\)

$\displaystyle \lim_{{u}\to{\infty}}\tanh\frac{1-e^{-2u}}{1+e^{-2u}} =\tanh(1)$

 

[MarkFL]

$\displaystyle \lim_{{u}\to{\infty}}\frac{1-e^{-2u}}{1+e^{-2u}} =1$

Yes, and so what does this tell you about the terminal velocity?

 

[karush]

presume unchanging but how is that going to give ft/sec

 

[MarkFL]

presume unchanging but how is that going to give ft/sec

If the hyperbolic tangent factor goes to 1, then the terminal velocity $v_T$ must be:

\(\displaystyle v_T=\sqrt{\frac{mg}{k}}\)

Dimensional analysis shows that $k$ must have units of mass per length, and then the velocity function has units of length per time.