A body of mass M is falling from a very high building. The drag force is given by F=kv^2 where k is a constant and v its velocity. How soon will it reach terminal velocity and what will the distance traversed be equal to?

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A body of mass M is falling from a very high building. The drag force is given by F=kv^2 where k is a constant and v its velocity. How soon will it reach terminal velocity and what will the distance traversed be equal to?

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Answer 1 · 2017-12-09T13:35:31

$The terminal velocity is approached exponentially but never$ $"The terminal velocity is approached exponentially but never"$ $reached exactly.$ $"reached exactly."$

Explanation:

$t = (\frac{1}{2}) \sqrt{\frac{M}{k g}} ln ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{1 + \sqrt{1 - exp (- 2 k \frac{h}{M})}}{1 - \sqrt{1 - exp (- 2 k \frac{h}{M})}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$ $t = (1/2)sqrt(M/(kg)) ln[(1+sqrt(1-exp(-2 k h/M)))/(1-sqrt(1-exp(-2kh/M)))]$
$v_{max} = sqrt(M g/k)$
$v = sqrt[ (M g/k) (1 - exp(-2 k h / M))]$

$So if v -> v_{max} , " " then " " h -> oo , " and so " t -> oo$

These formulas are obtained if one solves the differential equation

$M {dv}/dt = M g - k v^2$

Note that

$h = "height (measured downwards) fallen in meter"$

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A body of mass M is falling from a very high building. The drag force is given by F=kv^2 where k is a constant and v its velocity. How soon will it reach terminal velocity and what will the distance traversed be equal to?

1 Answer

Explanation:

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