Question

λλA particle falls from rest in a medium in which the resistance is λv^2 per unit mass, v being velocity of the particle at time t. Prove that the distance fallen in time to is 1/λ[Incos(t(√gλ))],where g is the acceleration due to gravity

Answer 1

Let $m$ be the mass of the particle. The weight of the particle is $mg$ and the drag force is $m\lambda v^2$ .

The equation of motion of the particle is

$m\frac{dv}{dt}=mg-m\lambda v^2\\ \Rightarrow \frac{dv}{dt}=g-\lambda v^2\\ \Rightarrow \frac{dv}{dt}=g(1-\frac{\lambda }{g}v^2)\\ \Rightarrow \frac{dv}{1-\frac{\lambda }{g}v^2}=gdt$

Integrating both sides,

$\int \frac{dv}{1-\frac{\lambda }{g}v^2}=g\int dt+C_1\\ \int \frac{dv}{1-\frac{\lambda }{g}v^2}=gt+C_1\ \quad ......(1)$

where $C_1$ is the constant of integration.

Let $z=\sqrt{\frac{\lambda}{g}}\ v\Rightarrow dz=\sqrt{\frac{\lambda}{g}}\ dv$

From (1), $\sqrt{\frac{g}{\lambda}}\int \frac{dz}{1-z^2}=gt+C_1$

$\therefore \sqrt{\frac{g}{\lambda}}\ \tanh^{-1}(z)=gt+C_1$ since $\int \frac{dz}{1-z^2}=\tanh^{-1}(z)$

Substituting $z=\sqrt{\frac{\lambda}{g}}\ v$ ,

$\sqrt{\frac{g}{\lambda}}\ \tanh^{-1}\left(\sqrt{\frac{\lambda}{g}}\ v\right)=gt+C_1\ \quad .......(2)$

At $t=0, v=0$

$\Rightarrow \sqrt{\frac{g}{\lambda}}\ \tanh^{-1}(0)=0+C_1\\ \Rightarrow C_1=0$ since $\tanh^{-1}(0)=0$

From (2),

$\sqrt{\frac{g}{\lambda}}\ \tanh^{-1}\left(\sqrt{\frac{\lambda}{g}}\ v\right)=gt\\ \Rightarrow \tanh^{-1}\left(\sqrt{\frac{\lambda}{g}}\ v\right)=gt\sqrt{\frac{\lambda}{g}}\\ \Rightarrow \tanh^{-1}\left(\sqrt{\frac{\lambda}{g}}\ v\right)=\sqrt{g\lambda }\ t\\ \Rightarrow \sqrt{\frac{\lambda}{g}}\ v=\tanh(\sqrt{g\lambda }\ t)\\ \Rightarrow v=\sqrt{\frac{g}{\lambda}}\tanh(\sqrt{g\lambda }\ t)\\ \Rightarrow \frac{dx}{dt}=\sqrt{\frac{g}{\lambda}}\tanh(\sqrt{g\lambda }\ t)\\ \Rightarrow dx=\sqrt{\frac{g}{\lambda}}\tanh(\sqrt{g\lambda }\ t)dt\\$

Integrating,

$x=\sqrt{\frac{g}{\lambda}}\int \tanh(\sqrt{g\lambda }\ t)dt+C_2\ \quad .......(3)$

where $C_2$ is the integration constant.

Let $y=\sqrt{g\lambda }\ t\Rightarrow dy=\sqrt{g\lambda }\ dt$

From (3),

$x=\sqrt{\frac{g}{\lambda}}\ \frac{1}{\sqrt{g\lambda }}\int \tanh(y)dy+C_2\\ \Rightarrow x=\frac{1}{\lambda} \int \tanh(y)dy+C_2\\ \Rightarrow x=\frac{1}{\lambda} \ln|\cosh(y)|+C_2$ since $\int \tanh(y)dy=\ln |\cosh(y)|$

Substituting $y=\sqrt{g\lambda }\ t$ ,

$x=\frac{1}{\lambda} \ln|\cosh(\sqrt{g\lambda }\ t)|+C_2\ \quad .......(4)$

At $t=0, x=0$

$\Rightarrow 0=\frac{1}{\lambda} \ln|\cosh(0)|+C_2$

Since $\cosh(0)=1 \ and\ \ln(1)=0$

$\Rightarrow C_2=0$

From (4),

$x=\frac{1}{\lambda} \ln|\cosh(\sqrt{g\lambda }\ t)|$

Thus, the distance fallen in time $t$ is $\frac{1}{\lambda} \ln|\cosh(\sqrt{g\lambda }\ t)|$