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so I'm trying to derive an equation for the maximum height an object reaches after being vertically thrown upwards, where the two forces acting on it are gravity and quadratic air resistance, so that

$$ ma=-mg-kv^2. $$

I'm trying to "fill in the gaps" of the derivation from this text, p. 2 under "Alternative Method".

I can accept that you can regard a differential quotient as a fraction. But I can't really logic my way through these two crucial steps with rules of math, where

$$ dz=\frac{mv}{mg+kv^{2}} dv $$

becomes

$$ \int_{0}^{h} dz=-\int_{V}^0\frac{mv}{mg+kv^{2}} dv. $$

$z$=height at a given time

$v$=velocity at a given time

$V$=initial velocity (right after the object is thrown)

$h$=maximum height the object reaches

It makes sense that when the object has reached $h$, the velocity is $0$.

But why is it possible, in this specific case, to do this mathematically - to integrate each side of the equation with respect to a different variable? Is it possible to show that it's the same as integrating w.r.t. the same variable?

I know this might be a question for math.stackexchange, but I thought that there maybe was a some physics knowledge necessary to explain this.

I am familiar with integration by substitution, separation of variables, and the fact that the antiderivative of velocity w.r.t. time is position (and that the derivative of position is velocity).

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    $\begingroup$ Probably somewhat less rigorous, v and z are both functions of time, Like when integrating parametrically, write dz as (dz/dt)dt and dv as (dv/dt) dt and now we are integrating both with respect to time. $\endgroup$ Commented Mar 21, 2023 at 13:42

2 Answers 2

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You can find the solution of the differential equation of $u(t)$ then integrate it from $t_{1}$ to $t_{2}\rightarrow x(t) = \int_{t_{1}}^{t_{2}}u(t)dt$

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the ODE's are:

$$m\,\frac{dv(t)}{dt}=-m\,g-k\,v(t)^2\tag 1$$ $$\frac{dh(t)}{dt}=v(t)\tag 2$$

from equation (2) you obtain $~dt=\frac{dh}{v}~$

hench equation (1)

$$ m\,\frac{dv}{dh}\,v=-m\,g-k\,v^2\tag 3$$

you can solve this differential equation with the initial condition $~v(h=0)=v_0~$ and obtain $~v=v(h)~$. solving $~v(h)=0~$ for h , this is the result ,$~h_{\rm max}$, that you are looking for.

alternative you can separate the variable $~dv~$ and $~dh~$ you obtain (Eq. (3))

$$dh=-\frac{m\,v}{m\,g+k\,v^2}\,dv$$

integrating

$$\int_0^{h_{\rm max}}dh=-\int_{v_0}^0\frac{m\,v}{m\,g+k\,v^2}\,dv$$

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  • $\begingroup$ I don't know if it's much to ask, but can you show how you would solve (3) and how the last equation is separating the variables? Like in symbols and such? And how can I show that those boundaries are valid? $\endgroup$ Commented Mar 21, 2023 at 17:35
  • $\begingroup$ If you solve the Eq. (3) you need initial condition $~v(h=0)=0~$ this is the one boundary, the second boundary is $~v(h_{max})=0~$ $\endgroup$
    – Eli
    Commented Mar 21, 2023 at 17:40

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