Question

The equation of motion of a body falling under gravity is given by $\frac{dv}{dt}=g-\frac{g}{{\lambda }^2}{v}^2$.The Velocity as a function of time is? given that $v=0$ $t=0$ .

• A. $v={\lambda }^{2}tan{h}^{-1}\frac{gt}{\lambda }$
• B. $v=\lambda cot{h}^{-1}\frac{gt}{\lambda }$
• C. $v=\lambda tan{h}^{-1}\frac{gt}{\lambda }$
• D. $v=\lambda co{t}^2{h}^{-1}\frac{gt}{\lambda }$

Answer 1

The correct option is C $v=\lambda tan{h}^{-1}\frac{gt}{\lambda }$

We have $\frac{dv}{dt}=g-k{v}^2$

Now taking $k=\frac{g}{{\lambda }^2}$ for convenience

$\frac{dv}{dt}=g-\frac{g}{{\lambda }^2}{v}^2=\frac{g}{{\lambda }^2}({\lambda }^2-v^2)$

This is a differential equation of variable separable type.

$\therefore \frac{dv}{{\lambda }^2-v^2}=\frac{g}{{\lambda }^2}=\frac{g}{{\lambda }^2}dt$

By integration, $\frac{1}{2\lambda }\log \left(\frac{\lambda +n}{\lambda -v}\right)=\frac{g}{{\lambda }^2}t+c$

$\therefore \frac{1}{\lambda }\tan .\frac{1}{2}\log \left(\frac{1+v/\lambda }{1-v/\lambda }\right)=\frac{g}{{\lambda }^2}t+C$

Since $\tan h^{-1}x=\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)$

$\frac{1}{\lambda }\tan h^{-1}\frac{v}{\lambda }=\frac{g}{{\lambda }^2}t+c$

But by data, when $t=0,v=0$ $\therefore c=0$

$\therefore \frac{1}{\lambda }\tan h^{-1}\frac{v}{\lambda }=\frac{g}{{\lambda }^2}t$ $\therefore \tan h^{-1}\frac{v}{\lambda }=\frac{gt}{\lambda }$

$v=\lambda tan{h}^{-1}\frac{gt}{\lambda }$