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v_0=100e^{\frac{-v_0}{100}}\\

\ln {v_0}=\ln{e^{\frac{-v_0}{100}}}^{100}\\

[/tex]

the answer is [tex]v_0=56.7[/tex]

how to find [tex]v_0[/tex]

??

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- Thread starter transgalactic
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- #1

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- 0

v_0=100e^{\frac{-v_0}{100}}\\

\ln {v_0}=\ln{e^{\frac{-v_0}{100}}}^{100}\\

[/tex]

the answer is [tex]v_0=56.7[/tex]

how to find [tex]v_0[/tex]

??

- #2

Mark44

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Shooting Star

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v_0=100e^{\frac{-v_0}{100}}\\

\ln {v_0}=\ln{e^{\frac{-v_0}{100}}}^{100}\\

[/tex]

the answer is [tex]v_0=56.7[/tex]

how to find [tex]v_0[/tex]

??

This is one way of doing it.

[tex]

v_0=100e^{\frac{-v_0}{100}}.

[/tex]

Putting [itex]

{\textstyle{{{\rm v}_{\rm 0} } \over {100}}} = x, x = e^{ - x} \Rightarrow x = 1 - x + x^2 /2,

[/itex] retaining up to the second order term after expanding [itex]e^{-x}.[/itex]

Now solve for x to get the approximate value. Justify why you are neglecting one value.

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