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    Calculate the limit of a falling object (with air-resistance) using L'Hopital's rule.

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    This problem has been particularly confusing to me:

    "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=[(mg)/(c)][1-e^([-ct]/m)]

    where g is the acceleration due to gravity and c is a positive constant.
    (a)Calculate the limit as t approaches infinity of v. What's the meaning of this limit?
    (b)For fixed t, use L'Hospital's rule to calculate the limit as m approaches infinity of v. What can you conclude about the speed of a very heavy falling object?"

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    Solution Preview

    (a)
    v=[(mg)/(c)][1-e^([-ct]/m)]
    when t approaches to infinity, v=mg/c. This velocity is called terminal velocity which means when air resistance is considered, the ...

    Solution Summary

    The limit of a falling object (with air-resistance) is calculated using L'Hopital's rule.

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