I am searching for a solution for my problem concerning the trajectory of projectiles. I have the following formulas to calculate the $x$, $y$-coordinates at any given time:
\begin{align*} x &= (\text{velocity} \times \cos \alpha \times \text{time})+ \left( \frac{\text{wind}}{2} \times \text{time}^2 \right) \tag{1}\\ y &= (\text{velocity} \times \sin \alpha \times \text{time})+ \left( \frac{\text{gravity}}{2} \times \text{time}^2 \right)\tag{2} \end{align*}
This works perfectly good, now I would like to calculate from a certain point $(x,y)$ with a fixed angle $\alpha$. The required velocity,
$$\text{velocity} = \frac{x}{\cos \alpha} \times \left[ \frac{\text{gravity}}{2(x\tan \alpha - y)} \right]\tag{3}$$
This works if I set wind to zero, but as soon as it is greater/smaller, it doesn't work anymore. What am I missing in the last formula to get the right result with wind?
$\require{cancel}$ your idea is very good, although practically is impossible, we want to obtain $v$ straightforward from (1) and (2), we get $$t^2=\frac{2(x-vt\cos\alpha)}{w}~~~,~~~t^2=\frac{2(y-vt\sin\alpha)}{g}$$ $$\frac{x-vt\cos\alpha}{w}=\frac{y-vt\sin\alpha}{g}$$ where $w$ is wind, thus $$t=\frac{yw-xg}{v(w\sin\alpha-g\cos\alpha)}\tag{4}$$ substitute (4) in (1): $$2x=\cancel{v}\cos\alpha\frac{yw-xg}{\cancel{v}(w\sin\alpha-g\cos\alpha)}+\frac12w\left(\frac{yw-xg}{v(w\sin\alpha-g\cos\alpha)}\right)^2$$ Then $$v^2=\left(\frac{yw-xg}{w\sin\alpha-g\cos\alpha}\right)^2\left(\frac{w(w\sin\alpha-g\cos\alpha)}{4x(w\sin\alpha-g\cos\alpha)-2w\cos\alpha(yw-xg)}\right)\color{blue}{\dfrac{m}{s}}$$