ode — models of motion

$$F = -mg - rv$$

Using Newton’s second law

$$m frac{dv}{dt} = -mg - rv, quad or quad frac{dv}{dt} = -g - frac{r}{m}v.$$

Separate variables

$$frac{dv}{g + rv/m} = -dt$$

Integrate for v

$$v(t) = Ce^{-rt/m} - frac{mg}{r}$$

C is a constant

Terminal velocity

$$lim_{n to infty} v(t) = - frac{mg}{r}$$

For displacement

$$frac{dx}{dt} = v = Ce^{-rt/m} - frac{mg}{r}$$

Solve for it

$$x = -frac{mC}{r}e^{frac{-rt}{m}} - frac{mgt}{r} + A$$

A is another constant

$$m frac{dv}{dt} = -mg - k|v|v, quad or quad frac{dv}{dt} = -g - frac{k}{m}|v|v.$$

When v < 0,

$$frac{dv}{dt} = -g + frac{k}{m} v^2$$

Solve for it, we get

$$v(t) = - sqrt{frac{mg}{k}} frac{1 - Ae^{-2t sqrt{kg/m}}}{1 + Ae^{-2t sqrt{kg/m}}}$$

Finding the displacement

$$a = frac{dv}{dt} = frac{dv}{dx} cdot frac{dx}{dt} = frac{dv}{dx} cdot v$$

Substitute it into $frac{dv}{dt} = -g - frac{r}{m}|v|v$

$$v frac{dv}{dx} = -g - frac{k}{m}|v|v$$

How to solve?

If v>0

$$v frac{dv}{dx} = -g - frac{k}{m}v^2 = - frac{mg + kv^2}{m}$$

Separate variables

$$frac{vdv}{mg + kv^2} = -frac{dx}{m}$$

At $t = 0$, $x(0) = 0$, $v(0) = v_0$. At later $t = T$, $x(T) = x_{max}$, $.v(T) = 0$. Using definite integration

$$int_{v_0}^0 frac{vdv}{mg + kv^2} = - int_{0}^{x_{max}} frac{dx}{m}$$

Solve for $x_{max}$

$$x_{max} = frac{m}{2k} ln left( 1 + frac{kv_0^2}{mg} right)$$