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Q13E

Expert-verifiedFound in: Page 116

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**When the velocity v of an object is very large, the magnitude of the force due to air resistance is proportional to v ^{2} with the force acting in opposition to the motion of the object. A shell of mass 3 kg is shot upward from the ground with an initial velocity of 500 m/sec. If the magnitude of the force due to air resistance is 0.1v^{2}, when will the shell reach its maximum height above the ground? What is the maximum height?**

- The shell reaches at the maximum height in
**2.69 sec** - The maximum height is
**101.9248 m**.

Apply the formula for thermal speed

${\mathrm{v}}_{\mathrm{t}}=\sqrt{\frac{\mathrm{mg}}{\mathrm{b}}}\phantom{\rule{0ex}{0ex}}{\mathrm{v}}_{\mathrm{t}}=\sqrt{\frac{3(9.81)}{0.1}}\phantom{\rule{0ex}{0ex}}=17.15\phantom{\rule{0ex}{0ex}}$ (m = 3, g = 9.81, b = 0.1)

$\mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{mg}-{\mathrm{bv}}^{2}\phantom{\rule{0ex}{0ex}}-\frac{\mathrm{m}}{\mathrm{b}}\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{mg}}{\mathrm{b}}+{\mathrm{v}}^{2}\phantom{\rule{0ex}{0ex}}-\frac{\mathrm{m}}{\mathrm{b}}\frac{\mathrm{dv}}{\mathrm{dt}}={{\mathrm{v}}^{2}}_{\mathrm{t}}+{\mathrm{v}}^{2}\phantom{\rule{0ex}{0ex}}\int \frac{1}{{{\mathrm{v}}^{2}}_{\mathrm{t}}+{\mathrm{v}}^{2}}\mathrm{dv}=\int \frac{-\mathrm{b}}{\mathrm{m}}\mathrm{dt}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{tan}}^{-1}\frac{\mathrm{v}}{{\mathrm{v}}_{\mathrm{t}}}}{{\mathrm{v}}_{\mathrm{t}}}=\frac{-\mathrm{bt}}{\mathrm{m}}+\mathrm{c}\phantom{\rule{0ex}{0ex}}$

When v = 500 m/sec and t = 0 then

$\mathrm{c}=\frac{{\mathrm{tan}}^{-1}\frac{500}{{\mathrm{v}}_{\mathrm{t}}}}{{\mathrm{v}}_{\mathrm{t}}}$

$\frac{{\mathrm{tan}}^{-1}\frac{\mathrm{v}}{{\mathrm{v}}_{\mathrm{t}}}}{{\mathrm{v}}_{\mathrm{t}}}=\frac{-\mathrm{bt}}{\mathrm{m}}+\frac{{\mathrm{tan}}^{-1}\frac{500}{{\mathrm{v}}_{\mathrm{t}}}}{{\mathrm{v}}_{\mathrm{t}}}\phantom{\rule{0ex}{0ex}}{\mathrm{tan}}^{-1}\frac{\mathrm{v}}{{\mathrm{v}}_{\mathrm{t}}}=\frac{-{\mathrm{btv}}_{\mathrm{t}}}{\mathrm{m}}+{\mathrm{tan}}^{-1}\frac{500}{{\mathrm{v}}_{\mathrm{t}}}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{v}}{{\mathrm{v}}_{\mathrm{t}}}=\mathrm{tan}(\frac{-{\mathrm{btv}}_{\mathrm{t}}}{\mathrm{m}}+{\mathrm{tan}}^{-1}\frac{500}{{\mathrm{v}}_{\mathrm{t}}})\phantom{\rule{0ex}{0ex}}\mathrm{v}={\mathrm{v}}_{\mathrm{t}}.\mathrm{tan}(\frac{-{\mathrm{btv}}_{\mathrm{t}}}{\mathrm{m}}+{\mathrm{tan}}^{-1}\frac{500}{{\mathrm{v}}_{\mathrm{t}}})$

The shell reaches at maximum height when v=0 and =17.15,

Then

$0=17.15\mathrm{tan}(\frac{-0.1(17.15)\mathrm{t}}{3}+{\mathrm{tan}}^{-1}\frac{500}{17.15})\phantom{\rule{0ex}{0ex}}\frac{-0.1(17.15)\mathrm{t}}{3}={\mathrm{tan}}^{-1}\frac{500}{17.15}\phantom{\rule{0ex}{0ex}}\mathrm{t}=\frac{{\mathrm{tan}}^{-1}\frac{500}{17.15}}{0.57}\phantom{\rule{0ex}{0ex}}\mathrm{t}=2.69\mathrm{sec}\phantom{\rule{0ex}{0ex}}$

**Hence, the shell reaches at the maximum height in 2.69 sec **

$\mathrm{v}={\mathrm{v}}_{\mathrm{t}}.\mathrm{tan}(\frac{-{\mathrm{btv}}_{\mathrm{t}}}{\mathrm{m}}+{\mathrm{tan}}^{-1}\frac{500}{{\mathrm{v}}_{\mathrm{t}}})\phantom{\rule{0ex}{0ex}}\mathrm{v}=17.15\mathrm{tan}(-0.57\mathrm{t}+{\mathrm{tan}}^{-1}\frac{500}{17.15})\left(\text{b = 0.1, m = 3,t = 2.69}\right)\phantom{\rule{0ex}{0ex}}\frac{\mathrm{dx}}{\mathrm{dt}}=17.15\mathrm{tan}(-0.57\mathrm{t}+1.537)\phantom{\rule{0ex}{0ex}}\mathrm{dx}=17.15\mathrm{tan}(-0.57\mathrm{t}+1.537)\mathrm{dt}\phantom{\rule{0ex}{0ex}}\mathrm{x}=17.15(\frac{100\mathrm{ln}(\mathrm{cos}\frac{570\mathrm{t}-1537}{1000})}{57})+\mathrm{c}\phantom{\rule{0ex}{0ex}}$

When x=0 the value of c=-101.925.

$\mathrm{x}\left(\mathrm{t}\right)=17.15(\frac{100\mathrm{ln}(\mathrm{cos}\frac{570\mathrm{t}-1537}{1000})}{57})+101.925$

Put t=2.69 then

$\mathrm{x}\left(\mathrm{t}\right)=17.15(\frac{100\mathrm{ln}(\mathrm{cos}\frac{570(2.69)-1537}{1000})}{57})+101.925-0.0002+101.925\phantom{\rule{0ex}{0ex}}\mathrm{x}\left(\mathrm{t}\right)=101.9248\text{}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

**Hence, the maximum height is 101.9248 m.**

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