I have a slight problem (actually quite huge) with this question however I am also not 100% confident in my procedure and need a second or third opinion.
A particle is thrown vertically upwards with a v(velocity)=u m/s in a medium with a resistance of the square of velocity (r=v2). Determine an expression for the velocity as it returns to its point of projection (g, k, and u only).
So (I am new to this forum and do not know how to set out the work as neatly as I should be, so I am skipping the first step with vectors):
v . dv/dx=u-g-v2
(let integral sign = ~ (haha... this is terrible))
At x=0, v=u
(Using log laws (Oh god what a mess))
Alright, so, here is my problem. I can (and have) figured out the max height reached by subbing when v=0 xmax is reached and then start again from the top but instead of having mu-mg-mv2=ma, I'd have mg+mu-mv2=ma but; using what working I have been doing; I would have found an expression for x and not for v. Also, I realize that I can use dv/dt instead of v . dv/dx but I cannot see how this would achieve the wanted answer.
In short, I do not know where to go from here, am I right for finding the xmax or should I have been using dv/dt at the start, or something I haven't even considered?
(Directing me to a tutorial on how to make neat problems would help too)