# Speed of skydiver

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• Mar 19th 2011, 08:29 PM
chengbin
Speed of skydiver
Hi. I have trouble integrating the following problem.

The equation is basically $m\frac{dv}{dt} = mg - kv^2$ with the condition when t = 0, v = 50m/s

I know this is a separable DE, so rearranaging it becomes

$\int \frac{mdv}{mg-kv^2} = \int dt$

Now I have trouble integrating the left side. Trig substitution doesn't work, so it seems like it'll end up being logs. Any help? Thanks.
• Mar 19th 2011, 08:33 PM
Prove It
Factorise the denominator as a difference of two squares, then apply partial fractions.
• Mar 19th 2011, 08:49 PM
topsquark
Or note that
$\displaystyle \int \frac{dx}{1 - x^2} = tanh^{-1}(x) + C$

-Dan
• Mar 20th 2011, 08:34 AM
chengbin
I think I'm doing this wrong, because the values I can when I substitute time is really wrong.

$\int \frac{mdv}{mg-bv^2} = \int dt$

$\frac {\sqrt m \coth^{-1} (\frac {\sqrt b v}{\sqrt {gm}}}{\sqrt {bg}}= t + c$

When t = 0, v = 52m/s

$c = \frac {\sqrt m \coth^{-1} (\frac {52\sqrt b}{\sqrt {gm}}}{\sqrt {bg}}$

Substituting this back

$\frac {\sqrt m \coth^{-1} (\frac {\sqrt b v}{\sqrt {gm}}-\sqrt m \coth^{-1} (\frac {52\sqrt b}{\sqrt {gm}}}{\sqrt {bg}} = t$

Solving for v

$v = \frac{\sqrt {gm} \coth (\coth^{-1}(\frac {52\sqrt b}{\sqrt {gm}}) + \frac {t\sqrt {bg}}{\sqrt m}) }{\sqrt b}$

Let
g = 9.8
m = 140
b = 110

When I anything other than t = 0 as the value for t, the value of v gets really close to 3.53, which is the value of $\frac {\sqrt {gm}}{\sqrt b}$

What did I do wrong?
• Mar 20th 2011, 11:54 AM
topsquark
Quote:

Originally Posted by chengbin
I think I'm doing this wrong, because the values I can when I substitute time is really wrong.

$\displaystyle \int \frac{mdv}{mg-bv^2} = \int dt$

$\displaystyle \frac {\sqrt m \coth^{-1} (\frac {\sqrt b v}{\sqrt {gm}}}{\sqrt {bg}}= t + c$

You can use coth^(-1) instead of tanh^(-1) but note that
$\displaystyle coth^{-1}(x) = tanh^{-1} \left ( \frac{1}{x} \right )$

(By the way, if you put \displaystyle just after the [tex] tag it'll make the LaTeX statements easier to read.)

The solution to the integral above is
$\displaystyle \int \frac{m~dv}{mg - bv^2} = \sqrt{ \frac{m}{bg}}~tanh^{-1} \left ( v \cdot \sqrt{\frac{b}{mg}} \right )$

In terms of coth^(-1)
$\displaystyle \int \frac{m~dv}{mg - bv^2} = \sqrt{ \frac{m}{bg}}~tanh^{-1} \left ( v \cdot \sqrt{\frac{b}{mg}} \right ) = \sqrt{ \frac{m}{bg}}~coth^{-1} \left ( \frac{1}{v} \cdot \sqrt{\frac{mg}{b}} \right )$

which is why your numbers are off.

-Dan
• Mar 20th 2011, 05:45 PM
chengbin
Quote:

Originally Posted by topsquark
You can use coth^(-1) instead of tanh^(-1) but note that
$\displaystyle coth^{-1}(x) = tanh^{-1} \left ( \frac{1}{x} \right )$

(By the way, if you put \displaystyle just after the [tex] tag it'll make the LaTeX statements easier to read.)

The solution to the integral above is
$\displaystyle \int \frac{m~dv}{mg - bv^2} = \sqrt{ \frac{m}{bg}}~tanh^{-1} \left ( v \cdot \sqrt{\frac{b}{mg}} \right )$

In terms of coth^(-1)
$\displaystyle \int \frac{m~dv}{mg - bv^2} = \sqrt{ \frac{m}{bg}}~tanh^{-1} \left ( v \cdot \sqrt{\frac{b}{mg}} \right ) = \sqrt{ \frac{m}{bg}}~coth^{-1} \left ( \frac{1}{v} \cdot \sqrt{\frac{mg}{b}} \right )$

which is why your numbers are off.

-Dan

Hi. Thanks for the correction, but my numbers are still off.

This is what I get.

$\displaystyle \sqrt \frac {m}{bg} \coth^{-1} \bigg (\frac {1}{v} \cdot \sqrt \frac {mg}{b}\bigg ) = t + c$

$\displaystyle c = \sqrt \frac {m}{bg} \coth^{-1} \bigg (\frac {1}{52} \cdot \sqrt \frac {mg}{b}\bigg )$

$\displaystyle t = \sqrt \frac {m}{bg} \bigg (coth^{-1} \frac {1}{v}\cdot \sqrt \frac {mg} {b} \bigg } - \coth^{-1} \bigg (\frac {1}{52} \cdot \sqrt \frac {mg}{b}\bigg ) \bigg )$

$\displaystyle v = \frac {\sqrt \frac {mg}{b}}{\coth \bigg ( \coth^{-1}\bigg ( \frac {1} {52} \cdot \sqrt \frac {mg} {b} \bigg ) + \frac {t} {\sqrt \frac {m}{bg}}\bigg )}$

But when I substitute a value for t, the value quickly drops from 52 to < 1 from t = 0 to t = 1.
• Mar 20th 2011, 06:53 PM
topsquark
I don't know how to help any further. When I try for a value of the constant I keep betting imaginary numbers. (See here.) As all other numbers in the problem should be arbitrary I have to conclude your value of b = 110 is far too large. I have been trying to find typical values of b to compare, but have so far been unsuccessful. The integral is correct, but I can't get any further than that.

-Dan

I'm thinking this over again. Your object is already traveling faster than the terminal speed at t = 0. All problems of this type I have seen start with the object falling from rest. For some reason this is throwing the equation off. I'll look into it and get back to you when (if!) I solve it.
• Mar 20th 2011, 07:26 PM
Prove It
Using what I suggested...

$\displaystyle \frac{1}{mg - kv^2} = \frac{1}{(\sqrt{mg})^2 - (\sqrt{k}\,v)^2} = \frac{1}{(\sqrt{mg} - \sqrt{k}\,v)(\sqrt{mg}+\sqrt{k}\,v)}$.

Now applying Partial Fractions

$\displaystyle \frac{A}{\sqrt{mg} - \sqrt{k}\,v} + \frac{B}{\sqrt{mg} + \sqrt{k}\,v} = \frac{1}{(\sqrt{mg} - \sqrt{k}\,v)(\sqrt{mg} + \sqrt{k}\,v)}$

$\displaystyle \frac{A(\sqrt{mg} + \sqrt{k}\,v) + B(\sqrt{mg} - \sqrt{k}\,v)}{(\sqrt{mg} - \sqrt{k}\,v)(\sqrt{mg} + \sqrt{k}\,v)} = \frac{1}{(\sqrt{mg} - \sqrt{k}\,v)(\sqrt{mg} + \sqrt{k}\,v)}$

$\displaystyle A(\sqrt{mg} + \sqrt{k}\,v) + B(\sqrt{mg} - \sqrt{k}\,v) = 1$

$\displaystyle A\sqrt{mg} + A\sqrt{k}\,v + B\sqrt{mg} - B\sqrt{k}\,v = 1$

$\displaystyle (A - B)\sqrt{k}\,v + (A + B)\sqrt{mg} = 0v + 1$

$\displaystyle (A - B)\sqrt{k} = 0 \implies A - B = 0 \implies A = B$

and $\displaystyle (A + B)\sqrt{mg} = 1 \implies A + B = \frac{1}{\sqrt{mg}} \implies 2B = \frac{1}{\sqrt{mg}} \implies B = \frac{1}{2\sqrt{mg}} \implies A = \frac{1}{2\sqrt{mg}}$.

So $\displaystyle \frac{1}{mg - kv^2} = \frac{1}{2\sqrt{mg}(\sqrt{mg} - \sqrt{k}\,v)} + \frac{1}{2\sqrt{mg}(\sqrt{mg} + \sqrt{k}\,v)}$

Therefore

$\displaystyle \int{\frac{m}{mg - kv^2}\,dv} = \frac{m}{2\sqrt{mg}}\int{\frac{1}{\sqrt{mg} - \sqrt{k}\,v} + \frac{1}{\sqrt{mg} + \sqrt{k}\,v}\,dv}$

$\displaystyle = \frac{m}{2\sqrt{kmg}}}\int{\frac{\sqrt{k}}{\sqrt{m g} - \sqrt{k}\,v} + \frac{\sqrt{k}}{\sqrt{mg} + \sqrt{k}\,v}\,dv}$

$\displaystyle = \frac{m}{2\sqrt{kmg}}\left(-\ln{|\sqrt{mg} - \sqrt{k}\,v|} + \ln{|\sqrt{mg} + \sqrt{k}\,v|}\right) + C$

$\displaystyle = \frac{m}{2\sqrt{kmg}}\ln{\left|\frac{\sqrt{mg} + \sqrt{k}\,v}{\sqrt{mg} - \sqrt{k}\,v}\right|} + C$.

So the solution of your DE is

$\displaystyle t = \frac{m}{2\sqrt{kmg}}\ln{\left|\frac{\sqrt{mg} + \sqrt{k}\,v}{\sqrt{mg} - \sqrt{k}\,v}\right|} + C$, and since you know that when $\displaystyle t = 0, v = 50$...

$\displaystyle 0 = \frac{m}{2\sqrt{kmg}}\ln{\left|\frac{\sqrt{mg} + 50\sqrt{k}}{\sqrt{mg} - 50\sqrt{k}}\right|} + C$

$\displaystyle C = -\frac{m}{2\sqrt{kmg}}\ln{\left|\frac{\sqrt{mg} + 50\sqrt{k}}{\sqrt{mg} - 50\sqrt{k}}\right|}$.

Therefore $\displaystyle t = \frac{m}{2\sqrt{kmg}}\ln{\left|\frac{\sqrt{mg} + \sqrt{k}\,v}{\sqrt{mg} - \sqrt{k}\,v}\right|} - \frac{m}{2\sqrt{kmg}}\ln{\left|\frac{\sqrt{mg} + 50\sqrt{k}}{\sqrt{mg} - 50\sqrt{k}}\right|}$

$\displaystyle t = \frac{m}{2\sqrt{kmg}}\ln{\left|\frac{(\sqrt{mg} - 50\sqrt{k})(\sqrt{mg} + \sqrt{k}\,v)}{(\sqrt{mg} + 50\sqrt{k})(\sqrt{mg} - \sqrt{k}\,v)}\right|}$
• Mar 20th 2011, 07:53 PM
topsquark
I guess the different forms can be useful! Thank you for saving me hours of work on this. :)

(sighs) Now I'm going to have to figure out why one solution worked and the other didn't.

I'm going to bed.

-Dan
• Mar 21st 2011, 05:04 AM
chengbin
Thanks for all your hard work.

@topsquark

This is not a problem from a textbook. I'm trying to model someone falling, but has reached terminal velocity already, hence when t = 0, v = 50. I wanna know his speed 2 seconds after he releases his parachute at t = 0.

The value for k is from the drag equation, $\frac {1}{2}\rho v^2C_dA$

p is 1.22 (density of air)
Cd = 2 (large drag for a parachute)
A = 100m^2 (area of the parachute)

The fact you experts are having trouble with this is making me doubt if I'm doing this question correctly.
• Mar 21st 2011, 09:24 AM
topsquark
Quote:

Originally Posted by chengbin
Thanks for all your hard work.

@topsquark

This is not a problem from a textbook. I'm trying to model someone falling, but has reached terminal velocity already, hence when t = 0, v = 50. I wanna know his speed 2 seconds after he releases his parachute at t = 0.

The value for k is from the drag equation, $\frac {1}{2}\rho v^2C_dA$

p is 1.22 (density of air)
Cd = 2 (large drag for a parachute)
A = 100m^2 (area of the parachute)

The fact you experts are having trouble with this is making me doubt if I'm doing this question correctly.

I see now. Yes a parachute would have a rather large value of b. Solve Prove It's solution for v and plug in 2 s for t. You will find that the speed is drastically reduced.

-Dan
• Mar 21st 2011, 03:14 PM
chengbin
• Mar 21st 2011, 06:03 PM
Prove It
I don't see what's so strange about it...
• Mar 21st 2011, 06:05 PM
chengbin
Quote:

Originally Posted by Prove It
I don't see what's so strange about it...

It drops from 50 to 3.53 in an instant. I don't think a parachute slows you down from 50m/s to 3.53m/s in an instant. That's why there is a relatively high minimum altitude you need to deploy your parachute while skydiving.
• Mar 21st 2011, 06:10 PM
topsquark
Quote:

Originally Posted by chengbin
It drops from 50 to 3.53 in an instant. I don't think a parachute slows you down from 50m/s to 3.53m/s in an instant.

Ever used a parachute? Take a look at that b value. At a first glance the numbers look about right to me as well.

-Dan
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