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Q 3.4-1E

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Found in: Page 115

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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# An object of mass 5 kg is released from rest 1000 m above the ground and allowed to fall under the influence of gravity. Assuming the force due to air resistance is proportional to the velocity of the object with proportionality constant b = 50 N-sec/m, determine the equation of motion of the object. When will the object strike the ground?

• The equation of motion of the object is $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}{\mathbf{.}}{\mathbf{981}}{\mathbf{t}}{\mathbf{-}}{\mathbf{0}}{.}{0}\mathbf{981}\mathbf{\left(}\mathbf{1}\mathbf{-}{{\mathbf{e}}}^{\mathbf{-}\mathbf{10}\mathbf{t}}{\mathbf{\right)}}{}{\text{m}}$

• The time taken by the object is 1019 sec.
See the step by step solution

## Step 1: Find the equation of motion of an object

The given values are ${\text{m = 5, g = 9.81, v}}_{\text{0}}\text{= 0, b = 50,}$

The equation of motion is role="math" localid="1664170601861" $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{mgt}}{\mathbf{b}}{\mathbf{+}}\frac{\mathbf{m}}{\mathbf{b}}\left(\mathbf{v}\mathbf{-}\frac{\mathbf{mg}}{\mathbf{b}}\right)\left(\mathbf{1}\mathbf{-}{\mathbf{e}}^{\frac{\mathbf{-}\mathbf{bt}}{\mathbf{m}}}\right)$

Put all the given values

$\frac{5\left(9.81\right)\mathrm{t}}{50}+\frac{5}{50}\left(0-\frac{5\left(9.81\right)}{50}\right)\left(1-{\mathrm{e}}^{\frac{-50\mathrm{t}}{5}}\right)\phantom{\rule{0ex}{0ex}}\frac{\left(9.81\right)\mathrm{t}}{10}-\frac{\left(9.81\right)}{100}\right)\left(1-{\mathrm{e}}^{-10\mathrm{t}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{x}\left(\mathrm{t}\right)=\frac{\left(9.81\right)\mathrm{t}}{10}-\frac{\left(9.81\right)}{100}\right)\left(1-{\mathrm{e}}^{-10\mathrm{t}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{x}\left(\mathrm{t}\right)=0.981\mathrm{t}-0.0981\left(1-{\mathrm{e}}^{-10\mathrm{t}}\right)\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Hence the equation of motion is role="math" localid="1664170624317" $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}{\mathbf{.}}{\mathbf{981}}{\mathbf{t}}{\mathbf{-}}{\mathbf{0}}{\mathbf{.}}{\mathbf{0}}\mathbf{981}\mathbf{\left(}\mathbf{1}\mathbf{-}{{\mathbf{e}}}^{\mathbf{-}\mathbf{10}\mathbf{t}}{\mathbf{\right)}}{\mathbf{}}{\mathbf{\text{m}}}$

## Step 2: Finding when will the object strike the ground

$1000=0.981\mathrm{t}-0.0981\left(1-{\mathrm{e}}^{-10\mathrm{t}}\right)\phantom{\rule{0ex}{0ex}}1000=0.981\mathrm{t}-0.0981\left(Becausevalueofissosmallsoneglecting\right)\phantom{\rule{0ex}{0ex}}1000+0.0981=0.981\mathrm{t}\phantom{\rule{0ex}{0ex}}1000.0981=0.981\mathrm{t}\phantom{\rule{0ex}{0ex}}\mathrm{t}=1000.0981/0.981\phantom{\rule{0ex}{0ex}}\mathrm{t}=1019\mathrm{sec}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Hence, the object will strike the ground in 1019 sec.

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