# Frames of reference and drag

I'm reading Jaan Kalda's handout on kinematics. There is a problem whose solution I don't understand.

After being kicked by a footballer, a ball started to fly straight towards the goal at velocity v = 25 m/s making an angle α = arccos 0.8 with the horizontal. Due to side wind blowing at u = 10 m/s perpendicular the initial velocity of the ball, the ball had deviated from its initial course by s = 2 m by the time it reached the plane of the goal. Find the time that it took the ball to reach the plane of the goal, if the goal was situated at distance L = 32 m from the footballer.

Publicly available solution states that

What does $$t_{air}$$ represent? Why is it different from $$t$$? If the ball's lateral displacement isn't proportional to time, then why isn't $$u$$ something like $$u_{air}$$? Also how is the displacement of the moving(which is that?) frame $$s$$?

P.S. I've also seen the solution video on YouTube that states that the balls trajectory in the wind's reference frame is a straight line? How?

Let's choose the air rest frame coordinate system such that the footballer is at the origin at $$t=0$$, and the initial velocity has the components $$\boldsymbol v_0= v\cos \alpha\ \boldsymbol{e}_x - u\ \boldsymbol{e}_y + v\sin\alpha\ \boldsymbol{e}_z$$, where $$\alpha = \arccos 0.8$$, $$v=25\,\mathrm{m}/\mathrm{s}$$ the initial ball speed, and $$u=10\,\mathrm{m}/\mathrm{s}$$ the wind speed. As the initial velocity and the ball's trajectory are parallel (when viewed from above), the following two triangles are similar, and the equations $$\frac{d}{L}=\frac{u}{v\cos\alpha}\qquad\Leftrightarrow d= \frac{uL}{v\cos \alpha}$$ hold, i.e. in the air's rest frame, the ball reaches the goal's plane in the point with horizontal cooordinates $$\left(L, -\frac{uL}{v\cos\alpha}\right)$$.
In the rest frame of the goal (with the origin being where the footballer made his shot) this point has horizontal coordinates $$\left(L, ut-\frac{uL}{v\cos\alpha}\right).$$ As we know that the ball's displacement is $$s=2\,\mathrm{m}$$, this leads to $$s = ut-\frac{uL}{v\cos\alpha}\\\qquad\Leftrightarrow\qquad ut = \underbrace{\left(\frac{uL}{v\cos\alpha}\right)}_{\begin{array}{c}\text{lateral displacement}\\\text{in air rest frame}\end{array}}+ \underbrace{s}_{\begin{array}{c}\text{lateral displacement}\\\text{in the goal's rest frame}\end{array}}$$ or, equivalently, $$\boxed{t = \frac{L}{v\cos\alpha} + \frac{s}{u}.}$$