Thread: Huge Intergration Help Needed

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.

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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):
mu-mg-mv²=ma
v . dv/dx=u-g-v²
dx/dv=v/(u-g-v²)
(let integral sign = ~ (haha... this is terrible))
~dx=~(v/(u-g-v²) dv
x=1/2~2v/(u-g-v²) dv
Therefore: x=1/2(ln(u-g-v²))+C
At x=0, v=u
Therefore: 0=1/2(ln(u-g-u²))+C
C=-1/2(ln(u-g-u²))

x=1/2(ln(u-g-v²))-1/2(ln(u-g-u²))
(Using log laws (Oh god what a mess))
x=1/2(ln(u-g-v²)/(u-g-u²))

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Alright, so, here is my problem. I can (and have) figured out the max height reached by subbing when v=0 x_max 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 x_max 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)

Just a couple of points to begin. In your first line why have you put mu in and why have you put mv^2 for the resistance?

Because it is a particle projected vertically upwards at a velocity of mu as the starting velocity is u, not the roaming (or actual) velocity with respect to time. Here it is graphically: mu <----P----> mg+mv^2 only imagine mu facing upwards so that drag is facing downwards.

Or is that wrong in any way? Doesn't seem to be.

On the upward journey there are two forces acting on the particle. weight mg and resistance v^2, both vertically downwards.
Let x be the distance travelled vertically then the 'force=mass*acceleration equation is -mg-v^2=mvdv/dx

That doesn't help me in the slightest. I know my ma equation is correct for the work that I am doing atm. However all I need to know is if my working is correct and where I can go from there. Also if there is any way to work around having dv/dt for a.

Why are you so sure your ma equation is correct?

Because I have checked with fellow classmates, however I cannot confirm my procedure with them until a much later date as they are on leave.

mu is not a force acting on the particle.

ok, so, I see what your saying aaaaand it is as u is not the force acting on it at time t, it is a force applied at time 0. So, also considering your objection, my ma would equal u-mg-mv^2=ma... so... I cannot go anywhere from here, literally. If I divided by m all the way through the equation I would have two unknowns in my integration and that is just bad news and does not make any sense with the given question.

u isn't a force at all. There are only 2 forces acting, the weight and the resistance.

Let's go with the assumption that my ma is correct and continue with my question.

Your 'ma' is definitely wrong so we can't help you any further if you won't accept the fact.

alright then, what you are proposing is that my ma is u-mg-mv^2 right? So, show me how I can get either my dv/dt or my v . dv/dx because it is seemingly impossible to me.

For the upward motion the equation is -v^2-mg=mvdv/dx