Newtonian dynamics

Question

A particle of mass $m$ is projected vertically upwards with speed $v_0$. The resistance force is of magnitude $m$$\lambda$$v^2$. Show that the particle comes to rest after rising through a distance:

$h$ = $1/2\lambda$ $(ln(1+\lambda v_0^2/g)$

Answer 1

Using Newton's Law, $$-mg-m\lambda v^2=-ma=-mv\frac{dv}{dx}$$ $$\Rightarrow \int^0_{v_0}\frac {vdv}{g+\lambda v^2}=-\int^H_0 dx$$ $$\Rightarrow \frac{1}{2\lambda}[\ln(g+\lambda v^2)]^0_{v_0}=-H$$ $$\Rightarrow H=\frac{1}{2\lambda}\ln\left(\frac{g+\lambda v^2}{g}\right)$$