Falling object and the side wind I don't have a big knowledge on Physics so I'm sorry in advance if it doesn't make sense.. 
If a brick is falling from 100m tall-building when there's the side wind of 30m/s, is there any way I can find how far the brick would have traveled from the origin?
If there's any equation I can utilize please let me know. I'm from architecture and trying to find out how to protect pedestrians from the falling objects. Thank you very much!  
 A: There's no easy way but it is possible to 'model' what you're asking.
The brick in question has a velocity vector $\vec{v}$ made up of two components, that are completely independent of each other:


*

*A vertical one $v_y$:


For simplicity's sake we'll assume the no significant drag forces act in the $y$ direction.
If $t=0$ is the moment the brick starts falling, then (with $g=9.81\:\mathrm{ms^{-2}}$):
$$v_y=gt$$
and the distance travelled:
$$y=\frac12 gt^2$$
Using the height of the building $y=H$ we can then calculate the falling time as:
$$t_f=\sqrt{\frac{2H}{g}}$$


*A horizontal one $v_x$, caused by air drag:


For high Reynolds numbers, the brick will experience a horizontal force cause by the wind, modelled as:
$$F_D=\frac12 \rho u^2C_DA$$ 
Where $u$ is the sideways wind velocity and I refer to this reference for the other symbols' meaning. More precisely $u$ is the relative speed of wind $v_{wind}$ to horizontal velocity $v_x$. Assuming $v_{wind}\gg v_x$ then we can assume $u$ is constant then the force $F_D$ is also constant.
Using Newton:
$$F_D=ma_x$$
With $a_x$ the horizontal acceleration:
$$a_x=\frac{\mathrm{d}}{\mathrm{dt}}v_x$$
The distance $x$ travelled sideways is given by:
$$x=\frac12 a_xt^2$$
Because the horizontal and vertical movements are independent of each other, using $t_f$ would give us $x_f$, the distance travelled sideways during the fall:
$$\boxed{x_f\approx \frac{1}{2mg}\rho u^2 C_DAH}$$
Now, this is all very fine and dandy but finding a reliable value for the drag coefficient $C_D$ may prove the hardest part.
A: The usual, and only simple,  way of calculating the sideways displacement $d$ of a projectile in a cross wind $W$ is 
$$
d= W(T_{\rm air}-T_{\rm vacuum})
$$
Here $T_{\rm air} $ is time to target, (or ground in your case) taking into account air resistence, and $T_{\rm vacuum}$ is the time the projectile or brick would take to reach the target (or ground) if there were no air resistance.
You can easily derive this by considering the motion in the frame of the wind. You can also give a more complicated derivation by considering the sideways component of the  drag force and relating the backwards component of the drag to change in time to reach the target.  The hard thing, of course, is computing the drag.
