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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

# An object of mass 5 kg is given an initial downward velocity of 50 m/sec and then allowed to fall under the influence of gravity. Assume that the force in newtons due to air resistance is -10v, where v is the velocity of the object in m/sec. Determine the equation of motion of the object. If the object is initially 500 m above the ground, determine when the object will strike the ground.

• The equation of motion of the object is $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{4}{\mathbf{.}}{\mathbf{91}}{\mathbf{t}}{\mathbf{+}}{\mathbf{22}}\mathbf{.}\mathbf{55}\mathbf{\left(}\mathbf{1}\mathbf{-}{{\mathbf{e}}}^{\mathbf{-}\mathbf{2}\mathbf{t}}{\mathbf{\right)}}$
• The time takes the object hits the ground 97.24 sec.
See the step by step solution

## Step 1: Find the velocity

$\text{ma = W - bv}\left(\text{t}\right)\phantom{\rule{0ex}{0ex}}\text{ma = mg - bv}\left(\text{t}\right)$

Velocity,

$\mathrm{v}\left(\mathrm{t}\right)=\frac{\mathrm{mg}}{\mathrm{b}}+\left(\mathrm{v}-\frac{\mathrm{mg}}{\mathrm{b}}\right){\mathrm{e}}^{\frac{-\mathrm{bt}}{\mathrm{m}}}\phantom{\rule{0ex}{0ex}}\mathrm{v}\left(\mathrm{t}\right)=\frac{5\left(9.81\right)}{10}+\left(50-\frac{5\left(9.81\right)}{10}\right){\mathrm{e}}^{\frac{-10\mathrm{t}}{5}}\phantom{\rule{0ex}{0ex}}\mathrm{v}\left(\mathrm{t}\right)=4.905+45.095{\mathrm{e}}^{-2\mathrm{t}}\phantom{\rule{0ex}{0ex}}$

## Step 2: Find the equation of motion

$\mathrm{x}\left(\mathrm{t}\right)=\frac{\mathrm{mgt}}{\mathrm{b}}+\frac{\mathrm{m}}{\mathrm{b}}\left(\mathrm{v}-\frac{\mathrm{mg}}{\mathrm{b}}\right)\left(1-{\mathrm{e}}^{\frac{-\mathrm{bt}}{\mathrm{m}}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{x}\left(\mathrm{t}\right)=4.905\mathrm{t}+\frac{5}{10}\left(45.095\right)\left(1-{\mathrm{e}}^{-2\mathrm{t}}\right)\phantom{\rule{0ex}{0ex}}=4.91\mathrm{t}+22.55\left(1-{\mathrm{e}}^{-2\mathrm{t}}\right)\phantom{\rule{0ex}{0ex}}$

Hence the equation of motion is $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{4}{\mathbf{.}}{\mathbf{91}}{\mathbf{t}}{\mathbf{+}}{\mathbf{22}}\mathbf{.}\mathbf{55}\mathbf{\left(}\mathbf{1}\mathbf{-}{{\mathbf{e}}}^{\mathbf{-}\mathbf{2}\mathbf{t}}{\mathbf{\right)}}$

## Step 3: Find the value of t

Put x=500 and neglecting the exponential part

$500=4.91\mathrm{t}+22.55\phantom{\rule{0ex}{0ex}}\mathrm{t}=97.24\mathrm{sec}\phantom{\rule{0ex}{0ex}}$

Hence, the time takes the object hits the ground 97.24 sec.