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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?

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    Wind plays role of velocity not acceleration – 2017-02-02
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    Do you obtain (3) from (1) and (2).? – 2017-02-02
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    (3) is independent of the other 2 – 2017-02-02
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    How did you obtain. intuition.? It's dimension is wrong. – 2017-02-02
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    Part intuition / part searching on google for similar problems. What do you mean with the wrong dimension? – 2017-02-02
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    The dimension of left hand is $\dfrac{m}{s}$ but right $\dfrac{m}{s^2}$ – 2017-02-02

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$\require{cancel}$ 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 $$v^2=\left(\frac{yw-xg}{w\sin\alpha-g\cos\alpha}\right)^2\left(\frac{w(w\sin\alpha-g\cos\alpha)}{4x(w\sin\alpha-g\cos\alpha)-2w\cos\alpha(yw-xg)}\right)\color{blue}{\dfrac{m}{s}}$$

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    Holy man, thank you very much! I had quiet a bit to understand that I used the wrong dimensions, but as you show it, it is very understandable! Thank you so much. :) – 2017-02-02
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    I'm happy for this good idea you have posted, any way, you are welcome. – 2017-02-02