Integrating w.r.t. two different variables - a physics problem on the maximum height of a vertically thrown object

Question

I'm trying to derive an equation for the 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)

$h$=height the object reaches before falling

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 am familiar with the chain rule, 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).

Answer 1

I'm wondering if it's a typo.

$ma=-mg-kv^2$

$a=dv/dt$.

$\frac{mdv}{mg+kv^2}=-dt$

$\frac{(1/g)dv}{1+\frac{kv^2}{mg}}=-dt$

$u^2=kv^2/mg\implies du=\sqrt{k/mg}dv$

$\int \frac{\sqrt{mg/k}(1/g) }{1+u^2}du =\sqrt{mg/k}(1/g) \arctan{(v\sqrt{k/mg})} +C$

Regarding the Alternative Method, the chain method is used to perform a change of variables.

$v=dz/dt$

$a=\frac{dv}{dt}=\frac{dv}{dz}\frac{dz}{dt}=v\frac{dv}{dz}$

Plug that into the original differential equation and you get as the integrand a numerator as a multiple of the derivative of the denominator yielding an integral of the form $\int (1/u) du =\ln{u}+C$.

Answer 2

The bounds are set differently on each side because the variable of integration on each side is different; they are set so that when velocity is $V,0$, the height is $0,h$ respectively.

This is how we normally integrate. Suppose you had to solve for the displacement function $z(t)$ given a velocity function $v(t)$. You would note that

$${dz \over dt}=v(t)\iff z(t)-z(0)=\int_0^t v(t) dt\iff \int_{z(0)}^{z(t)}dz=\int_0^tv(t)dt.$$