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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?

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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.

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  • $\begingroup$ 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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