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The drag equation is: $F_D=\frac12 \rho C_d A v^2$, assuming constant values let's simplify to $F_D=Cv^2$

Assuming there are no other forces to consider, I'm really not sure where to go from here. I tried plugging it into $F=ma$, but now I just have an acceleration that is dependent on instantaneous velocity. What I would like is the ability to plug in an initial velocity along with the travel time, and get out the final velocity to see how much it's slowed due to drag. I'm having difficulty conceptualising this problem because at every step the amount by which the velocity is reduced depends on the velocity at that given moment, it's self referential. This leads me to believe calculus must be involved, but I'm not sure how to set up the problem. It seems like there should be an analytical solution.

edit: Since my post was marked as a duplicate of "Why do rain drops fall with a constant velocity?" - No, I am not asking about terminal velocity. Suppose we fire a spherical ball of known dimensions horizontally at a speed of 500m/s and let it fly for exactly 5 seconds, I am asking if there is an analytical way to solve exactly how much the horizontal speed will be reduced by (ignoring any other external forces). After that 5 second period, will it be going 250m/s, 300m/s, 350?? This would be easy for me to do were the force constant, I could just apply the kinematics formula: $v_f = v_i+at$. My predicament is that the acceleration is not constant, and I don't know how to account for that.

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  • $\begingroup$ See solution to your type of drag in the comments to this answer. See also Terminal velocity $\endgroup$
    – Roger V.
    Nov 9, 2022 at 8:27
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    $\begingroup$ this is called differential analysis $\endgroup$ Nov 9, 2022 at 8:38
  • $\begingroup$ The answer given by Roger Vadim on the linked duplicate actually mentions that the resulting equation is a differential equation, and it gives the solution. But it elides how this solution is arrived at, saying that it's done using "usual techniques", which I think is what the OP is unfamiliar with. So I'm voting to reopen, though I suspect that we could find an actual duplicate elsewhere on this site. $\endgroup$ Nov 9, 2022 at 15:05
  • $\begingroup$ @MichaelSeifert the comment that was pointed out in Roger Vadim's comment spells out (in somewhat suspect notation) the separation of variables and resulting integral in the $v^2$ drag case. $\endgroup$
    – NickD
    Nov 9, 2022 at 17:53

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As you have figured out, assuming that the velocity $v$ of the object is positive, the velocity of the object satisfies $$ m \frac{dv}{dt} = -C v^2 \quad \Rightarrow \quad \frac{dv}{dt} = -\alpha v^2, $$ where $\alpha \equiv C/m$. This is what's called an ordinary differential equation (ODE), since it relates the function $v(t)$ to its derivative with respect to $t$. The solution will be a function $v(t)$ which satisfies this relationship at every point and (in addition) has the correct initial velocity (i.e., $v(0) = v_0$ for whatever $v_0$ the object starts with.

There is no general method for finding exact analytical solutions to ODEs, but there are several methods that have been developed. Typically (at least in the USA), these techniques form a semester-long course in mathematics at the second-year level; describing all of them here would be impossible. However, I will point out that this particular equation is what's called a first-order equation, and it is separable; and the Wikipedia page linked above includes a summary of general results that includes this case, so you might try tackling it from there.

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  • $\begingroup$ First question, why did you put a negative sign on the right hand side? $\endgroup$
    – Scott
    Nov 10, 2022 at 9:47
  • $\begingroup$ Next, I did try this actually, solving the ode using separation of variables, here was my method: $\int\frac{1}{v^2}\:dv=\int-A\:dt\\$, ==> $-\frac{1}{v}+C_1=-At+C_2\\$, ==> $\frac{1}{v}=At+C_3\\$, ==> $v=\frac{1}{At+C_3}$, I'm not sure what I've accomplished here though. I've seemingly ended up with a linear function, where I know the reduction in velocity can not be linear. I also have an unknown constant. $\endgroup$
    – Scott
    Nov 10, 2022 at 10:23
  • $\begingroup$ @Scott: The negative sign is there because a drag force is in the opposite direction of the motion; so if $v$ is positive, $dv/dt$ must be negative. Your answer looks correct to me, and it's not linear; linear would mean that (for example) $\Delta v$ between $t = 0$ and $t = 1$ would be the same as $\Delta v$ between $t = 1$ and $t=2$, but it's not hard to show that this is not the case. Finally, the constant $C_3$ is determined by the condition that $v(0) = v_0$; in this case, this implies that $C_3 = 1/v_0$. $\endgroup$ Nov 10, 2022 at 12:41

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