# Help

• Apr 21st 2006, 09:55 PM
Help
Actually, I have no idea how to do this problem :confused:
• Apr 22nd 2006, 12:30 AM
CaptainBlack
Quote:

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

Take a coordinate system with $y$ positive downwards, and $y=0$ corresponding to the parachutists initial position.

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

$
f=32m-8v
$

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

$
32m-8v=m \dot v
$

or in terms of $y$:

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

rearranging gives:

$
\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:

$
\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 $y=e^{\lambda t}$, then:

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

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

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

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

$
\dot y= 4m
$

so the PI is:

$
y(t)=4m t
$
,

so the general solution is:

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

Now at $t=0$ we have $y=0$ so $A+B=0$.
Also $\dot y =0$ gives $-\frac{8}{m}A+4m=0$, so the solution is:

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