How do you integrate velocity with respect to dy for an integration problem?

[pt176900]

What is the technique for integrating velocity with respect to dy?

In other words:
integral(dv^2/dt) dy [ from 0 to h ]

if anyone can help, that'd be super.. mkay

 

[HallsofIvy]

Without knowing, the relationship of velocity to y, no!

You don't even say, though I would guess it, the the velocity is the velocity of something in the y direction- that is, that v= dy/dt. d(v²)/dt= 2 v dv/dt.
We still can't integrate that with respect to y until we have some idea of the relationship between y and t.

 

[pt176900]

at time t = 0, y = 0
at some time t, y will be a maximum.

The problem reads: Find the speed of a bullet fired straight up when the resistive force is given by F(v) = -cv^2
so you have

1/2 mdv^2/dy = -mg - cv^2
mdv^2/dy = 2 (-mg -cv^2)

which is a seperable differential equation

mdv^2 = 2 (-mg - cv^2) dy

dv^2 = (-2g - 2cv^2/m) dy = -2 (g + cv^2/m) dy

so... this is where I am a little stuck
can you suggets where to proceed?

 

[HallsofIvy]

The whole point of a "separable equation" is to separate the variables!

Once you have md(v²)= 2 (-mg - cv²) dy

Rewrite it as $m\frac{d(v^2)}{cv^2+ mg}= -2ydy$ and integrate.

(I started to write d(v²) as 2v dv but then I recognized that you have only v² and not v itself in the integral. If you let y= v², the integral on the left is exactly the same as the integral of $m\frac{dy}{cy+mg}$.)