# Answer in Differential Equations

Let “m” be the mass of the particle. The weight of the particle is “mg” and the drag force is “mlambda v^2” .

The equation of motion of the particle is

“mfrac{dv}{dt}=mg-mlambda v^2\nRightarrow frac{dv}{dt}=g-lambda v^2\nRightarrow frac{dv}{dt}=g(1-frac{lambda }{g}v^2)\nRightarrow frac{dv}{1-frac{lambda }{g}v^2}=gdt”

Integrating both sides,

“int frac{dv}{1-frac{lambda }{g}v^2}=gint dt+C_1\nint 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}} vRightarrow 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}} vright)=gt+C_1 quad …….(2)”

At “t=0, v=0”

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

From (2),

“sqrt{frac{g}{lambda}} tanh^{-1}left(sqrt{frac{lambda}{g}} vright)=gt\nRightarrow tanh^{-1}left(sqrt{frac{lambda}{g}} vright)=gtsqrt{frac{lambda}{g}}\nRightarrow tanh^{-1}left(sqrt{frac{lambda}{g}} vright)=sqrt{glambda } t\nRightarrow sqrt{frac{lambda}{g}} v=tanh(sqrt{glambda } t)\nRightarrow v=sqrt{frac{g}{lambda}}tanh(sqrt{glambda } t)\nRightarrow frac{dx}{dt}=sqrt{frac{g}{lambda}}tanh(sqrt{glambda } t)\nRightarrow dx=sqrt{frac{g}{lambda}}tanh(sqrt{glambda } t)dt\”

Integrating,

“x=sqrt{frac{g}{lambda}}int tanh(sqrt{glambda } t)dt+C_2 quad …….(3)”

where “C_2” is the integration constant.

Let “y=sqrt{glambda } tRightarrow dy=sqrt{glambda } dt”

From (3),

“x=sqrt{frac{g}{lambda}} frac{1}{sqrt{glambda }}int tanh(y)dy+C_2\nRightarrow x=frac{1}{lambda} int tanh(y)dy+C_2\nRightarrow x=frac{1}{lambda} ln|cosh(y)|+C_2” since “int tanh(y)dy=ln |cosh(y)|”

Substituting “y=sqrt{glambda } t” ,

“x=frac{1}{lambda} ln|cosh(sqrt{glambda } 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{glambda } t)|”

Thus, the distance fallen in time “t” is “frac{1}{lambda} ln|cosh(sqrt{glambda } t)|”

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