Answer to Question #137771 in Classical Mechanics for A

Question #137771
A particle is moving under gravity in a medium whose resistance is mkv^4. Find the motion where v is velocity
1
Expert's answer
2020-10-15T03:11:34-0400
ma=mgmkv4a=dvdt=gkv4ma=mg-mkv^4\\a=\frac{dv}{dt}=g-kv^4

c1+t=(log(g14k14v)+log(g14+k14v)+2tan1(k14g14v)4g34k14c_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}}}


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