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Q3.4-24E

Expert-verifiedFound in: Page 117

Book edition
9th

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

Pages
616 pages

ISBN
9780321977069

**Rocket Flight. A model rocket having initial mass mo kg is launched vertically from the ground. The rocket expels gas at a constant rate of a kg/sec and at a constant velocity of b m/sec relative to the rocket. Assume that the magnitude of the gravitational force is proportional to the mass with proportionality constant g. Because the mass is not constant, Newton’s second law leads to the equation (mo - αt) dv/dt - αβ = -g(m0 – αt), where v = dx/dt is the velocity of the rocket, x is its height above the ground, and m0 - αt is the mass of the rocket at t sec after launch. If the initial velocity is zero, solve the above equation to determine the velocity of the rocket and its height above ground for 0≤t<m0/α.**

- The velocity is $\mathbf{v}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}{\beta}{}{\mathbf{ln}}\left|\frac{{\mathbf{m}}_{\mathbf{o}}\mathbf{-}\alpha \mathbf{t}}{{\mathbf{m}}_{\mathbf{o}}}\right|{\mathbf{-}}{\mathbf{gt}}$

** **

- The height of the rocket is

$\mathbf{(}{\mathbf{m}}_{\mathbf{o}}\mathbf{-}\alpha \mathbf{t}\mathbf{)}\frac{\mathbf{dv}}{\mathbf{dt}}\mathbf{-}\alpha \beta \mathbf{=}\mathbf{-}\mathbf{g}\mathbf{(}{\mathbf{m}}_{\mathbf{o}}\mathbf{-}\alpha \mathbf{t}\mathbf{)}$

Since, the initial velocity is zero so v(0) = 0

$\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\alpha \beta}{{\mathrm{m}}_{\mathrm{o}}-\alpha \mathrm{t}}-\mathrm{g}\phantom{\rule{0ex}{0ex}}\int \mathrm{dv}=\int \frac{\alpha \beta}{{\mathrm{m}}_{\mathrm{o}}-\alpha \mathrm{t}}-\mathrm{g}\mathrm{dt}\left(\text{Integrating on both sides}\right)\phantom{\rule{0ex}{0ex}}\mathrm{v}\left(\mathrm{t}\right)=-\mathrm{\beta}\mathrm{ln}\left|{\mathrm{m}}_{\mathrm{o}}-\mathrm{\alpha t}\right|-\mathrm{gt}+\mathrm{C}\phantom{\rule{0ex}{0ex}}$

Thus, ${m}_{o}-\alpha t>0,\text{}t<{m}_{o}/\alpha \text{and}v\left(0\right)=0\text{}then\text{}C=-\beta \mathrm{ln}\left|{\mathrm{m}}_{\mathrm{o}}\right|$

$\mathrm{v}\left(\mathrm{t}\right)=-\beta \mathrm{ln}\left|{\mathrm{m}}_{\mathrm{o}}-\alpha \mathrm{t}\right|-\mathrm{gt}+-\beta \mathrm{ln}\left|{\mathrm{m}}_{\mathrm{o}}\right|\phantom{\rule{0ex}{0ex}}\mathrm{v}\left(\mathrm{t}\right)=-\beta \mathrm{ln}\left|\frac{{\mathrm{m}}_{\mathrm{o}}-\alpha \mathrm{t}}{{\mathrm{m}}_{\mathrm{o}}}\right|-\mathrm{gt}\phantom{\rule{0ex}{0ex}}$

**Hence, the velocity is $\mathbf{v}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}{\beta}{}{\mathbf{ln}}\left|\frac{{\mathbf{m}}_{\mathbf{o}}\mathbf{-}\alpha \mathbf{t}}{{\mathbf{m}}_{\mathbf{o}}}\right|{\mathbf{-}}{\mathbf{gt}}$**

$\mathrm{x}\left(\mathrm{t}\right)=\int (-\beta \mathrm{ln}\left|\frac{{\mathrm{m}}_{\mathrm{o}}-\alpha \mathrm{t}}{{\mathrm{m}}_{\mathrm{o}}}\right|-\mathrm{gt})\mathrm{dt}\phantom{\rule{0ex}{0ex}}\mathrm{x}\left(\mathrm{t}\right)=-\frac{\beta ({\mathrm{m}}_{\mathrm{o}}-\alpha \mathrm{t})}{{\mathrm{m}}_{\mathrm{o}}}\left[\mathrm{ln}\left|\frac{{\mathrm{m}}_{\mathrm{o}}-\alpha \mathrm{t}}{{\mathrm{m}}_{\mathrm{o}}}\right|-1\right]+\mathrm{C}$

When x(0) = 0 then C = - ,then

$\mathrm{x}\left(\mathrm{t}\right)=\frac{\beta (\alpha \mathrm{t}-{\mathrm{m}}_{\mathrm{o}})}{{\mathrm{m}}_{\mathrm{o}}}\left[\mathrm{ln}\left|\frac{{\mathrm{m}}_{\mathrm{o}}-\alpha \mathrm{t}}{{\mathrm{m}}_{\mathrm{o}}}\right|-1\right]-\beta $

**Hence, the height of the rocket is role="math" localid="1664218019166" $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\beta \mathbf{(}\alpha \mathbf{t}\mathbf{-}{\mathbf{m}}_{\mathbf{o}}\mathbf{)}}{{\mathbf{m}}_{\mathbf{o}}}\left[\mathbf{ln}\left|\frac{{\mathbf{m}}_{\mathbf{o}}\mathbf{-}\alpha \mathbf{t}}{{\mathbf{m}}_{\mathbf{o}}}\right|\mathbf{-}\mathbf{1}\right]{\mathbf{-}}{\beta}$ .**

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