Q13E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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?

Step-by-Step Solution

Answer

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

Step 1: Find the thermal speed of the object

Apply the formula for thermal speed 

 $$\begin{gathered} {\mathrm{v}}_{\mathrm{t}}=\sqrt{\frac{\mathrm{mg}}{\mathrm{b}}} \\ {\mathrm{v}}_{\mathrm{t}}=\sqrt{\frac{3(9.81)}{0.1}} \\ =17.15 \end{gathered}$$             (m = 3, g = 9.81, b = 0.1)

Step 2: Find the value of velocity

$$\begin{gathered} \mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{mg}-{\mathrm{bv}}^2 \\ -\frac{\mathrm{m}}{\mathrm{b}}\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{mg}}{\mathrm{b}}+{\mathrm{v}}^2 \\ -\frac{\mathrm{m}}{\mathrm{b}}\frac{\mathrm{dv}}{\mathrm{dt}}={{\mathrm{v}}^2}_{\mathrm{t}}+{\mathrm{v}}^2 \\ \int \frac{1}{{{\mathrm{v}}^2}_{\mathrm{t}}+{\mathrm{v}}^2}\mathrm{dv}=\int \frac{-\mathrm{b}}{\mathrm{m}}\mathrm{dt} \\ \frac{{\tan }^{-1}\frac{\mathrm{v}}{{\mathrm{v}}_{\mathrm{t}}}}{{\mathrm{v}}_{\mathrm{t}}}=\frac{-\mathrm{bt}}{\mathrm{m}}+\mathrm{c} \end{gathered}$$

When v = 500 m/sec and t = 0 then 

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

$$\begin{gathered} \frac{{\tan }^{-1}\frac{\mathrm{v}}{{\mathrm{v}}_{\mathrm{t}}}}{{\mathrm{v}}_{\mathrm{t}}}=\frac{-\mathrm{bt}}{\mathrm{m}}+\frac{{\tan }^{-1}\frac{500}{{\mathrm{v}}_{\mathrm{t}}}}{{\mathrm{v}}_{\mathrm{t}}} \\ {\tan }^{-1}\frac{\mathrm{v}}{{\mathrm{v}}_{\mathrm{t}}}=\frac{-{\mathrm{btv}}_{\mathrm{t}}}{\mathrm{m}}+{\tan }^{-1}\frac{500}{{\mathrm{v}}_{\mathrm{t}}} \\ \frac{\mathrm{v}}{{\mathrm{v}}_{\mathrm{t}}}=\tan (\frac{-{\mathrm{btv}}_{\mathrm{t}}}{\mathrm{m}}+{\tan }^{-1}\frac{500}{{\mathrm{v}}_{\mathrm{t}}}) \\ \mathrm{v}={\mathrm{v}}_{\mathrm{t}}.\tan (\frac{-{\mathrm{btv}}_{\mathrm{t}}}{\mathrm{m}}+{\tan }^{-1}\frac{500}{{\mathrm{v}}_{\mathrm{t}}}) \end{gathered}$$

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

Then

 $$\begin{gathered} 0=17.15\tan (\frac{-0.1(17.15)\mathrm{t}}{3}+{\tan }^{-1}\frac{500}{17.15}) \\ \frac{-0.1(17.15)\mathrm{t}}{3}={\tan }^{-1}\frac{500}{17.15} \\ \mathrm{t}=\frac{{\tan }^{-1}\frac{500}{17.15}}{0.57} \\ \mathrm{t}=2.69 \sec \end{gathered}$$

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

Step 3: Find the maximum height

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

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

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

Put t=2.69 then 

$$\begin{gathered} \mathrm{x}\left(\mathrm{t}\right)=17.15(\frac{100\ln (\cos \frac{570(2.69)-1537}{1000})}{57})+101.925-0.0002+101.925 \\ \mathrm{x}\left(\mathrm{t}\right)=101.9248\mathrm{m} \end{gathered}$$

Hence, the maximum height is 101.9248 m.