Finding the final velocity with quadratic drag

[CrosisBH]

Homework Statement

You shoot a particle straight up with a velocity $$v_0 < v_{t}$$ (terminal velocity) in a medium (like air) where the drag fore is proportional to the square of the velocity, show that the particle lands with a speed given by $$v_f = \frac{v_0 v_t}{\sqrt{v_0^2 + v_t^2}}$$ (Hint, change $$\frac{dv}{dt}$$ to $$v\frac{dv}{dx}$$)

Relevant Equations

$$\vec{F} = m\vec{a}$$
$$v_t = \sqrt{\frac{mg}{c}}$$

I chose coordinates where down is positive. So the force going up is $$F_{up} = mg - cv^2$$
$$a = g + \frac{c}{m}v^2$$
$$a = g + \frac{c}{m}v^2$$
$$a = g \left(1 + \frac{v^2}{v_t^2}\right)$$
$$a = \frac{dv}{dt} = v\frac{dv}{dy} = g \left(1 + \frac{v^2}{v_t^2}\right)$$
I used normal separation of variables to solve this. I noticed that the velocity starts at v_0 and goes to a velocity 0 because gravity is opposite the velocity. And that point where it hits 0 the y will be at the max, so I did so.
$$\int_{v_0}^{0} \frac{v}{\left(1 + \frac{v^2}{v_t^2}\right)}dv = \int_{0}^{y_{max}}gdy $$
I evaluated to get
$$y_{max} = \frac{-v_t^2\ln\left(\frac{v_t^2 + v_0^2}{v_t^2}\right)}{2g}$$
This dimensionally makes sense, and also physically makes sense because I chose positive in the down direction.

I used this information to solve the down case,
$$F_{down} = mg-cv^2$$
Using very similar algebra I reached here.
$$\int_{0}^{v_f} \frac{v}{\left(1 - \frac{v^2}{v_t^2}\right)}dv = \int_{y_{max}}^{0}gdy $$

As I'm writing this post I realized I just went circular here, and was getting a solution of $$v_t = iv_0$$, which obviously makes no sense. I just don't know how to proceed after finding the max y. Any help is appreciated.

 

[TSny]

Your work looks good to me. But I don't see why you say that you went in a circle. Did you actually carry out the final integration? I don't believe you will find that $v_t = iv_0$. You're looking for $v_f$.[I did notice a typographical error in your equation $F_{up} = mg - cv^2$ where the minus sign on the right should be a + sign, but you corrected that in the next line.]

 

[Chestermiller]

Homework Statement:: You shoot a particle straight up with a velocity $$v_0 < v_{t}$$ (terminal velocity) in a medium (like air) where the drag fore is proportional to the square of the velocity, show that the particle lands with a speed given by $$v_f = \frac{v_0 v_t}{\sqrt{v_0^2 + v_t^2}}$$ (Hint, change $$\frac{dv}{dt}$$ to $$v\frac{dv}{dx}$$)
Relevant Equations:: $$\vec{F} = m\vec{a}$$
$$v_t = \sqrt{\frac{mg}{c}}$$

I chose coordinates where down is positive. So the force going up is $$F_{up} = mg - cv^2$$
$$a = g + \frac{c}{m}v^2$$
$$a = g + \frac{c}{m}v^2$$
$$a = g \left(1 + \frac{v^2}{v_t^2}\right)$$
$$a = \frac{dv}{dt} = v\frac{dv}{dy} = g \left(1 + \frac{v^2}{v_t^2}\right)$$

There should be a minus sign in front of the mg. The equation should read: $$mv\frac{dv}{dy}=-mg-cv^2$$

 

[TSny]

There should be a minus sign in front of the mg. The equation should read: $$mv\frac{dv}{dy}=-mg-cv^2$$

For the flight upward, I think the signs on the right should be positive if the y-axis increases downward.
(For the flight upward, $v<0$, $dv >0$, and $dy < 0$)

 

[Chestermiller]

For the flight upward, I think the signs on the right should be positive if the y-axis increases downward.
(For the flight upward, $v<0$, $dv >0$, and $dy < 0$)

It doesn’t make sense (to me) to have y measured downward for the flight upward,

 

[TSny]

Either choice of the orientation of the y-axis is ok. For the flight upward, choosing the y-axis downward makes keeping track of the signs harder. ($y_{max}$ is then negative, $v_0$ is also negative, etc.) But, it works out. The relation $v_0 < v_t$ should then be written $|v_0| < v_t$, assuming $v_t$ is defined to be positive.

 

[CrosisBH]

Okay it turns out I was making a fatal error. These damn negative signs. The second differential equation yields
$$\frac{-v_t^2 \ln\left(\frac{v_t^2-v_f^2}{v_t^2}\right)}{2g} = -y_{max}$$. I forgot cancel negatives and dropped it when i subbed y_max. A lot of canceling yields:
$$\ln\left(\frac{v_t^2 - v_f^2}{v_t^2}\right) = - \ln\left(\frac{v_t^2+v_0^2}{v_t^2}\right)$$
Which with some algebra and a lot of manipulation can yield.
$$v_f = \frac{v_0 v_t}{\sqrt{v_t^2 + v_0^2}}$$
By combining fractions and stuff.

Also the reason I used down is positive so I can have mg be a positive quantity. I get really confused with what sign drag should have when writing down force equations, so I followed the simple example with no external force my professor used, where down was positive. Thank you everyone!