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Q5E

Expert-verifiedFound in: Page 115

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 $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{4}{\mathbf{.}}{\mathbf{91}}{\mathbf{t}}{\mathbf{+}}{\mathbf{22}}\mathbf{.}\mathbf{55}\mathbf{(}\mathbf{1}\mathbf{-}{{\mathbf{e}}}^{\mathbf{-}\mathbf{2}\mathbf{t}}{\mathbf{)}}$
- The time takes the object hits the ground
**97.24 sec**.

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

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

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

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

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.**

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