Question

A particle of mass m is projected vertically up with an initial velocity of Vo . If the force due to the friction of the air is directly proportional to its instantaneous velocity,
Calculate the velocity and position of the particle as a function of time?

Answer 1

As per the question,

Mass of the particle $=M$

Initial velocity of the particle $=V_o$

Let the particle is moving upwards, so force of friction due to air resistance will downwards.

$F=-mg-kv_o$

Where k is constant.

$t=\int_{v_o}^{v}\frac{mdv}{F(v)}$

$=-m\int_{v_o}^{v}\frac{ dv}{mg+kv}$

$=\frac{-m}{k}\ln(mg+kv)|_{vo}^v$

$=\frac{-m}{k}\ln(\frac{mg+kv_o}{mg+kv})$

$\Rightarrow \frac{-kt}{m}=\ln(\frac{mg+kv_o}{mg+kv})$

$\Rightarrow \frac{mg+kv_o}{mg+kv}=e^{\frac{-kt}{m}}$

$\Rightarrow v=(\frac{mg}{k}+v_o)e^{\frac{-k}{m}t}-\frac{mg}{k}$

$v=\frac{dx}{dt}$

Hence,

$dx=((\frac{mg}{k}+v_o)e^{\frac{-k}{m}t}-\frac{mg}{k})dt$

Now, taking the integration of both side of the above equation,

$\int_{x_o}^{x} dx=\int_o^t ((\frac{mg}{k}+v_o)e^{\frac{-k}{m}t}-\frac{mg}{k})dt$

$\Rightarrow x=x_o-\frac{mgt}{k}+(\frac{m^2 g}{k^2}+\frac{mv_o}{k})[1-e^{\frac{-kt}{m}}]$