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.

  • $\begingroup$ Would the surface of the ball have an effect? golf ball cf tennis ball cf cricket ball or similar? $\endgroup$
    – Solar Mike
    Nov 25 '21 at 20:29
  • $\begingroup$ @SolarMike, yes, the surface of the ball would change the drag. Some machines shoot with a pitch, which would affect drag too. $\endgroup$
    – kamran
    Nov 25 '21 at 20:33

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