q.
A heavy particle is projected at speed $U$ at an angle $\alpha$ to the horizontal. The particle is subject to air resistance which is experimentally found to vary proportionally to the square of the speed. Show that
$$\vec{\dot{v}} = -\frac{g}{V^2}\lvert\vec{v}\rvert\vec{v} - g\vec{j},$$
where $V$ is the terminal velocity of the particle. If $\alpha = \frac{\pi}{2}$ (so that the particle is projected directly upwards), find the maxium height reached and the time taken to reach it. What is the speed of the particle when it returns to the horizontal?
I'm on the part where "If $\alpha = \frac{\pi}{2}$ ..."
seeing as $\alpha = \frac{\pi}{2}$ it would be acceptable to think of motion in one dimension, so from the "show that" equation I get $$\ddot r = \dot v = \dfrac{-g}{V^2}v^2 - g $$
Now I'm curious here, can I just integrate w.r.t to t? Usually in vector form v is a function of t, and I'm asuming it is here aswell, but is this allowed:
$$\ddot r = \dfrac{-g}{V^2}v^2 - g $$ $$\dot r = \left ( \dfrac{-g}{V^2}v^2 - g \right )t + C$$
If so, why? If not, why not? Furthermore, I get a complex solution to the terminal velocity, what does this mean - As the terminal velocity can't be 0, what can we conclude from that?
edit: solving $\dot v = -kv^2 - g$ where $k = \dfrac{g}{V^2}$ I get
$\dfrac{dv}{kv^2+g} = -dt$ and solving this I get $v = \dfrac{\sqrt{g}}{\sqrt{k}} arctan(C\sqrt{k}\sqrt{g} - \sqrt{k}\sqrt{g}t)$ where C is a constant, integrating this again I get $r = \dfrac{1}{k}\log (sec(c\sqrt{k}\sqrt{g} - \sqrt{k}\sqrt{g}t)) + C_1$ then for $v = 0$ I get $ t = C$ so $r = C_1$, fro r = 0 I get $t = \dfrac{C\sqrt{k}\sqrt{g} - arcsec(e^{-Ck})}{\sqrt{k}\sqrt{g}}$ so $v = \dfrac{\sqrt{g}}{\sqrt{k}}tan(arcsec(e^{-Ck})) = \dfrac{\sqrt{g}}{\sqrt{k}}\sqrt{1 - \frac{1}{e^{-2Ck}}}e^{-Ck}$
