2-3-26 - k = 0 and see that the term goes to v To evaluate...

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2.3.26 We are considering a body with mass m , initial velocity v 0 and a resistance of k | v | . Resistance is a force. By Newton, | F | = m | a | , so | a air resistance | = | F | /m . Then we have v 0 = a air resistance + a gravity Resistance opposes the direction of movement, so a air resistance = - kv/m . Thus v 0 = - kv m - g This is a separable equation, so we can solve it to ﬁnd v = - mg k + ± mg k + v 0 ² e - kt/m Suppose air resistance goes to zero. That is, consider what happens as k 0. Rearranging terms we see lim k 0 v = lim k 0 ³ mg ( - 1 + e - kt/m ) k + v 0 e - kt/m ´ The second term in the limit is a continuous function, so we can just plug in
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Unformatted text preview: k = 0 and see that the term goes to v . To evaluate the limit of the ﬁrst term, we use L’Hˆ opital’s rule: lim k → mg (-1 + e-kt/m ) k = mg lim k →-t m e-kt/m 1 =-gt Putting it all together, we have that lim k → v = v-gt If we instead considered m → 0, note that e-kt/m → 0 for t > 0, so we immediately see lim m → v = 0...
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This note was uploaded on 07/17/2008 for the course MATH 415 taught by Professor Costin during the Fall '07 term at Ohio State.

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