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  1. #1
    Junior Member
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    Help

    Actually, I have no idea how to do this problem
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    Last edited by CaptainBlack; April 21st 2006 at 10:55 PM. Reason: to fit image to scree size
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by askmemath
    Actually, I have no idea how to do this problem
    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:

    <br />
f=32m-8v<br />

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

    <br />
32m-8v=m \dot v<br />

    or in terms of y:

    <br />
m \ddot y + 8 \dot y=32m<br />
,

    rearranging gives:

    <br />
\ddot y + \frac{8}{m} \dot y=32<br />
.

    Now this may be solved in the usual manner, the solution the sum of a
    solution to the homogeneous equation:

    <br />
\ddot y + \frac{8}{m} \dot y=0<br />

    and a particular integral of the original equation.

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

    <br />
\lambda ^2 +\frac{8}{m} \lambda =0<br />
,

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

    <br />
y(t)=Ae^{-\frac{8}{m}t}+B<br />
.

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

    <br />
\dot y= 4m<br />

    so the PI is:

    <br />
y(t)=4m t<br />
,

    so the general solution is:

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

    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:

    <br />
y(t)=\frac{m^2}{2} e^{-\frac{8}{m}t}+4m t -\frac{m^2}{2}<br />
.

    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)
    Last edited by CaptainBlack; April 21st 2006 at 11:33 PM.
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