So I’ve been trying to work out the range of a ball that has been projected by two spinning discs please when accounting for drag-
This is what I got so far: $V=V\cos\theta$, $T=2V\sin\theta/g$ , and $Range=V*T,$ thus, $Range= v^2\sin(2\theta)/g,$ but this doesn’t account for drag?
Where do I go from here please and is the above even correct please?
when there is air resistance the horizontal component of velocity is not constant anymore and it decays.
Let's set the origin at the launch point with the X, Z axes, and C as positive constant friction or drag of the air. The equation of motion:
$$ m\,\frac{d{\bf v}}{dt} = m\,{\bf g} - c\,{\bf v},$$
$$ V = ( V_X, V_Z) \quad g=(0,-g),v_t =terminla \ velocity=mg/c$$
$$ \frac{dv_x}{dt}= - g\,\frac{v_x}{v_t},$$
$$ \frac{d v_z}{dt}= - g\left(1+\frac{v_z}{v_t}\right).$$
Integrating X component
$$ \int_{v_{x\,0}}^v\frac{dv_x}{v_x} = - \frac{g}{v_t}\,t, $$
Wher $ v_{x\,0} = v_0\,\cos\theta$ is the $x$-component of the launch velocity. therefore
$$\ln\left( \frac{V_X}{V_{X_0}}\right)= \frac{-g}{V_t}t, \ then $$
$$V_x=v_0cos\theta*e^{-gt/v_t} $$.
The x component will decay exponentially.
Applying the vertical forces and integrating we get:
$$V_Z=V_0sin\theta*e^{-gt/V_t}-V_t* \left(1-e^{-gt/V_T} \right)$$
Among other things, the above implies that if the ball stays in the air much longer than the order of $,V_t/g \ $it will fall vertically. I know this answer can use some details and graphs, I come back and finish them.
Aero drag is proportional to v^2, which means there is no algebraic solution. So you either approximate, by linearising it over the range of speed of interest (this would get ugly quick), or use a numerical solution (which is what is usually done).
