Need help with fluid mechanics problem set by lecturer :[

[Keval]

$$\Delta t \left(gt-\frac{k}{m}v^2 \right)= \Delta v$$
If your trying to work out the change in velocity how can you use v^2 on the left when that's wht your trying to work out for this part.

and your saying once i get that sorted i use $\frac{\Delta v }{\Delta t }=s$

 

[nvn]

Keval: I'm currently not understanding the explanation either. Not to mention several other things, velocity is changing nonlinearly. Therefore, you need to use differentials, not differences.

Anyway, it is good you found an air drag equation, but I currently think you should keep researching and looking for a different equation. In particular, the first equation http://en.wikipedia.org/wiki/Drag_(physics)#Velocity_of_a_falling_object". Remember, v(t) is dx/dt, so change v(t) in that equation to dx/dt. Then multiply both sides of the equation by dt. Next, integrate both sides. I.e., integrate one side from zero to x, and integrate the other side from zero to t.

By the way, I have not checked your Cd = 0.47 value. Ensure you are using the Cd value your teacher wants you to use, and/or a Cd value roughly applicable to your velocity range.

 

[Keval]

Keval: I'm currently not understanding the explanation either. Not to mention several other things, velocity is changing nonlinearly. Therefore, you need to use differentials, not differences.

Anyway, it is good you found an air drag equation, but I currently think you should keep researching and looking for a different equation. In particular, the first equation http://en.wikipedia.org/wiki/Drag_(physics)#Velocity_of_a_falling_object". Remember, v(t) is dx/dt, so change v(t) in that equation to dx/dt. Then multiply both sides of the equation by dt. Next, integrate both sides. I.e., integrate one side from zero to x, and integrate the other side from zero to t.

By the way, I have not checked your Cd = 0.47 value. Ensure you are using the Cd value your teacher wants you to use, and/or a Cd value roughly applicable to your velocity range.

Hey just got back from uni and saw thiss could you use some latex to show me? I'm sort of confused, am still reading up on it thought

 

[nvn]

You can write the latex, if you wish. Use the hints. Show your work; and then someone might check your math.

 

[Keval]

okay here's trying this new method i get lost pretty early :\

$F=ma$

$mg-F_d=ma$ using $F_d=-\frac{1}{2}\rho A C_d v^2$

$mg-kv^2=ma$ using $k=-\frac{1}{2}\rho A C_d$

$mg-k(\frac{ds}{dt})^2=m\frac{dv}{dt}$

$mg-k\frac{d^2s}{dt^2}=m\frac{dv}{dt}$

now what? o.o, I'm not sure if i shoud be using v=dx/dt, or v=ds/dt.

And I've never done 2nd order de's if that's what I've ended up with sofar so helping hand please

 

[viscousflow]

Hello Keval,

I looked at nvn's equations and how he's trying to help you. I think he's trying to say that Drag is equal to some constant times velocity, not velocity squared.

 

[nvn]

Keval: Your post 40 is not what I advised in post 37. Read what I wrote more carefully. viscousflow, I didn't think I was saying what you said. I thought I was saying, velocity is a function of time.