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Expert-verified Found in: Page 116 ### 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 # When the velocity v of an object is very large, the magnitude of the force due to air resistance is proportional to v2 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 0.1v2, 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.
See the step by step solution

## Step 1: Find the thermal speed of the object

Apply the formula for thermal speed

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

## Step 2: Find the value of velocity

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

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

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

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

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

Then

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

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

## Step 3: Find the maximum height

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

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

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

Put t=2.69 then

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

Hence, the maximum height is 101.9248 m. ### Want to see more solutions like these? 