Question

The distance $x$ descended by a parachuter is satisfied by the differential equation, ${\left(\frac{dx}{dt}\right)}^2=k^2[1-e^{-2gx/k^2}]$.where ,$k,g$ are constants.If $x=0$ at $t=0$ then which of the following is the correct solution?

• A. $x=\frac{k^2}{g}\log \cos \frac{gt}{k}$
• B. $x=\frac{k^2}{g}\log \sin h\frac{gt}{k}$
• C. $x=\frac{k^2}{g}\log \cos h\frac{gt}{k}$
• D. $x=\frac{k^2}{g}\log \tan h\frac{gt}{k}$

Answer 1

The correct option is C $x=\frac{k^2}{g}\log \cos h\frac{gt}{k}$

We have $\frac{dx}{dt}b=k\sqrt{1-e^{-2gx/k^2}}$

$\therefore \frac{dx}{\sqrt{1-e^{-2gx/k^2}}}=kdt$

Let $\sqrt{1-e^{-2gx/k^2}}=u$ $\therefore 1-u^2{e}^{-2gx/k^2}=u^2$

$\therefore e^{-2gx/k^2}\cdot \frac{g}{k^2}\cdot dx=udu$ $\therefore (1-u^2)\frac{g}{k^2}dx=udu$

$\therefore dx=\frac{k^2}{g}\cdot \frac{u}{1-u^2}du$

Hence, from (1), we get

$\frac{k^2}{g}\cdot \frac{u}{1-u^2}\frac{1}{u}du=kdt$ $\therefore \frac{g}{g}\cdot \frac{du}{1-u^2}=dt+c$

By integration, $\frac{k}{g}\cdot \frac{1}{2}\log \left(\frac{1+u}{1-u}\right)=t+c$

But $\frac{1}{2}\log \left(\frac{1+u}{1-u}\right)=\tan h^{-1}u$ $\therefore \frac{k}{g}\tan h^{-1}u=t+c$

But by data when $t=0,x=0$ and hence $u=0$ $\therefore c=0$

Therefore, $\frac{k}{g}\tan h^{-1}u=t$ $\therefore \frac{k}{g}\tan h^{-1}u=\frac{gt}{k}$

$\Rightarrow u=\tan h\frac{gt}{k}$ $\therefore u^2=\tan h^2\left(\frac{gt}{k}\right)$

$\Rightarrow 1-e^{-2gx/k^2}=\tan h^2\left(\frac{gt}{k}\right)$

$\Rightarrow e^{-2gx/k^2}=\tan h^2\left(\frac{gt}{k}\right)=\sec h^2\left(\frac{gt}{k}\right)$

$\Rightarrow e^{2gx/k^2}=\cos h^2\left(\frac{gt}{k}\right)$ $\Rightarrow \frac{2gx}{k^2}=2\log \cos h\left(\frac{gt}{k}\right)$

$\Rightarrow x=\frac{k^2}{g}\log \cos h\left(\frac{gt}{k}\right)$.