Forest UCM PnCP QuadAirRes

Consider a ball falling under the influence of gravity and a frictional force that is proportion to its velocity squared

$\sum \vec{F}_{ext} = mg -bv^2 = m \frac{dv}{dt}$

Find the fall distance

Here is a trick to convert the integral over time to one over distance so you don't need to integrate twice as inthe previous example

$\frac{dv}{dt} = \frac{dv}{dy}\frac{dy}{dt} = v\frac{dv}{dy}$

The integral becomes

$mg -bv^2 = m v\frac{dv}{dy}$

$\int_{y_i}^{y_f} dy = \int_{v_i}^{v_f} m \frac{vdv}{\left ( mg -bv^2 \right ) }$

$y = \int_{v_i}^{v_f} \frac{vdv}{\left ( g -\frac{b}{m}v^2 \right ) }$

let $u = g -\frac{b}{m}v^2$

then $du = -2\frac{b}{m}v dv$

$y =\int_{v_i}^{v_f} \frac{-m}{2b} \frac{du}{u } = \frac{b}{m} \int_{v_f}^{v_i} \ln {g -\frac{b}{m}v^2} =$

$y =\frac{m}{2b} \ln \left ( \frac {g -\frac{b}{m}v_i^2}{g -\frac{b}{m}v_f^2} \right )$

Forest_UCM_PnCP#quadratic_friction