# Help

• Apr 21st 2006, 08:55 PM
Help
Actually, I have no idea how to do this problem :confused:
• Apr 21st 2006, 11:30 PM
CaptainBlack
Quote:

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

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

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:

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