Answer to Question #121375 in Classical Mechanics for physics

Question

Force acting on an object falling from a height close to the earth's surface is given as
F = -mg + bv ^ 2
The first term is the force of gravity, and the second is the friction force and b is a constant, and v is the velocity. The initial conditions are given as x0 = v0 = t0 = 0.
Find the position and velocity of the particle as a function of time. All intermediate actions should be clearly written.
Find the approximate values of position and velocity functions for small and large values of time t. All intermediate procedures must be written clearly.

Answer 1

according to Newton’s second law:

$mg-b\dot{x}^2 = m\ddot{x}\\ \ddot{x}+\frac{b}{m}\dot{x}^2 = g$

find the complementary solution:

$\ddot{x}+\frac{b}{m}\dot{x} = 0\\ \dot{x}=y\\ \dot{y}+\frac{b}{m}y = 0\\ y(t)=c_1e^{-\frac{b}{m}t}\\ x(t)=c_1e^{-\frac{b}{m}t}+c_2$

and then general solution is:

$x(t)=c_1e^{-\frac{b}{m}t}+\frac{mgt}{b}+c_2\\ v=\dot{x} = \frac{-bc_1}{m}e^{-\frac{b}{m}t}+\frac{mg}{b}\\ v(0)=\frac{-bc_1}{m}+\frac{mg}{b} = 0\\ c_1=\frac{m^2g}{b^2}\\ x(0) = \frac{m^2g}{b^2}+c_2=0\\ c_2 = -\frac{m^2g}{b^2}\\ x(t)=\frac{m^2g}{b^2}e^{-\frac{b}{m}t}+\frac{mgt}{b}-\frac{m^2g}{b^2}\\ v(t) = -\frac{mg}{b}e^{-\frac{b}{m}t}+\frac{mg}{b}\\ \begin{cases} x(t)\sim \frac{mgt}{b}\\ v(t)\sim \frac{mg}{b} \end{cases} , t\to\infty\\ \begin{cases} x(t)\sim \frac{m^2g}{b^2}e^{-\frac{b}{m}t}\\ v(t)\sim -\frac{mg}{b}e^{-\frac{b}{m}t} \end{cases} ,t\to0$