Answer to Question #137771 in Classical Mechanics for A

Question

A particle is moving under gravity in a medium whose resistance is mkv^4. Find the motion where v is velocity

Answer 1

$$ma=mg-mkv^4\\a=\frac{dv}{dt}=g-kv^4$$

$$c_1 + t =\frac{ (-\log(g^{\frac{1}{4}} - k^{\frac{1}{4}}v) + \log(g^{\frac{1}{4}} + k^{\frac{1}{4}} v) +2 \tan^{-1}{\left(\frac{k^{\frac{1}{4}}}{g^{\frac{1}{4}}}v\right)}}{4 g^{\frac{3}{4}} k^{\frac{1}{4}}}$$