I'm having trouble with a high school mathematics question. An object of mass $1kg$ falls from rest in a medium in which the resistance to motion is given by $r=kv^2$, where $k$ is a constant and $v$ is the speed.
Prove that the distance, $x$, fallen when the velocity is $v$ is given by: $$x=\frac{1}{2k}log_e\left(\frac{g}{g-kv^2}\right )$$
Thanks for any help!
This is not a high school calculation (I need to figure it out) The governing equation $$ \text{Resultant Force} = \text{Force due to gravity} -\text{resistive Force} $$ thus using newtons law of $$ \text{Force} = \text{mass}\times\text{acceleration} = ma $$ We derive the following $$ ma = m\frac{dv}{dt} = mg - kv^2 $$ knowing that $v = \dfrac{dx}{dt}$, where $x$ is the distance traveled, we find $$ m\dfrac{d^2x}{dt^2} = mg - k\left(\frac{dx}{dt}\right)^2 $$ Where $$ a = \dfrac{dv}{dt} = \dfrac{d}{dt}\frac{dx}{dt} $$ for mass $m = 1kg$ we find $$ \ddot{x} = g - k\dot{x}^2 $$ can we solve this (or even use this method for HS I do not know) $$ \ddot{x} = \dfrac{dx}{dt}\dfrac{d}{dx}\dot{x} = \frac{1}{2}\dfrac{d}{dx}\dot{x}^2 = g - k\dot{x}^2 $$re-arrange to find $$ \frac{1}{g-k\dot{x}^2}d(\dot{x}^2) = 2dx \implies -\frac{1}{k}\ln \left(g-k\dot{x}^2\right) = 2x + C $$ remember $\dot{x} = v$ we have $$ -\frac{1}{k}\ln \left(g-kv^2\right) = 2x + C $$ lets find $C$ we have $x(t=0) = v(t=0) = 0$ (my assuming that we start with x(0) = 0) $$ -\frac{1}{k}\ln g = C $$ so we have $$ 2x = -\frac{1}{k}\ln \left(g-kv^2\right) +-\frac{1}{k}\ln g = -\frac{1}{k}\ln\left(\frac{g-kv^2}{g}\right) = \frac{1}{k}\ln\left(\frac{g}{g-kv^2}\right) $$ so $$ x = \frac{1}{2k}\ln\left(\frac{g}{g-kv^2}\right) $$