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Problem 78 Hard Difficulty

If an object with mass $ m $ is dropped from rest, one model for its speed $ v $ after $ t $ seconds, taking air resistance into account, is
$$ v = \frac{mg}{c}(1 - e^{-ct/m}) $$
where $ g $ is the acceleration due to gravity and $ c $ is a positive constant. (In Chapter 9 we will be able to deduce this equation from the assumption that the air resistance is proportional to the speed of the object; $ c $ is the proportionality constant.)
(a) Calculate $ lim_{t\to \infty} v $. What is the meaning of this limit?
(b) For fixed $ t $, use l'Hospital's Rule to calculate $ lim_{c\to 0^+} v $. What can you conclude about the velocity of a falling object in a vacuum?

Answer

a)$\lim _{t \rightarrow \infty} \frac{m g}{c}\left(1-e^{-c t / m}\right)=\frac{m g}{c}(1-0)=\frac{m g}{c}$
b)$g t$

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Video Transcript

So for this problem we're having tea, go to infinity. Um and we see that when T goes to infinity, we end up getting things cancel out, and we get MG oversee that's the first portion. But then for part B, what we're gonna do is we're letting when we have the indeterminate form As C approaches zero from the right, what we end up having is to take the derivative of the top and bottom. So once we use the hotel's rule, we end up getting by treating see as a mhm variable, we end up getting the limit. Uh see approaches zero of energy times T over em of E. To the negative C. T. Over end. Then as we when we plug in uh C two B zero, we end up getting just G. T. As the final answer for that.

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