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About LinkBacksJust 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?
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):
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))
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)
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.
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.
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.
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