Actually, I have no idea how to do this problem :confused:

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- Apr 21st 2006, 08:55 PMaskmemathHelp
Actually, I have no idea how to do this problem :confused:

- Apr 21st 2006, 11:30 PMCaptainBlackQuote:

Originally Posted by**askmemath**

Now while the parachutist is moving downward the force on here is:

$\displaystyle

f=32m-8v

$

and Newton tells us that $\displaystyle f=m \dot v$ so:

$\displaystyle

32m-8v=m \dot v

$

or in terms of $\displaystyle y$:

$\displaystyle

m \ddot y + 8 \dot y=32m

$,

rearranging gives:

$\displaystyle

\ddot y + \frac{8}{m} \dot y=32

$.

Now this may be solved in the usual manner, the solution the sum of a

solution to the homogeneous equation:

$\displaystyle

\ddot y + \frac{8}{m} \dot y=0

$

and a particular integral of the original equation.

For the homogeneous equation, try a solution of the form $\displaystyle y=e^{\lambda t}$, then:

$\displaystyle

\lambda ^2 +\frac{8}{m} \lambda =0

$,

which has solutions $\displaystyle \lambda=0$ and $\displaystyle \lambda=-\frac{8}{m}$, so the general solution to HE is:

$\displaystyle

y(t)=Ae^{-\frac{8}{m}t}+B

$.

Now a particular integral of the original DE is the constant velocity solution:

$\displaystyle

\dot y= 4m

$

so the PI is:

$\displaystyle

y(t)=4m t

$,

so the general solution is:

$\displaystyle

y(t)=Ae^{-\frac{8}{m}t}+4m t +B

$

Now at $\displaystyle t=0$ we have $\displaystyle y=0$ so $\displaystyle A+B=0$.

Also $\displaystyle \dot y =0$ gives $\displaystyle -\frac{8}{m}A+4m=0$, so the solution is:

$\displaystyle

y(t)=\frac{m^2}{2} e^{-\frac{8}{m}t}+4m t -\frac{m^2}{2}

$.

Which is sufficient to answer the questions asked.

RonL

(you will need to check the algebra involved here, and that the assumption

that the parachutists velocity is always positive)