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

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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?


1 Answer 1


This problem nicely teaches the solver the importance of choosing a situation-appropriate coordinate system. Indeed, when looked at from the right perspective, the situation at hand turns into a simple problem of kinematics and does not require any complicated modelling of the ball's drag.

The key insight is that in the air's rest frame, the drag force is always directed opposite to the ball's instantaneous velocity vector. Since the only other force acting on the ball, gravity, merely changes the vertical component of the velocity vector, we can observe that there is no force that could change the velocity vector's horizontal orientation: in the rest frame of the air, the ball's trajectory is a straight line when viewed from above and is parallel to the initial velocity vector. Viewed from the side, we would see an air-restistance-modified parabola but we purposefully look away from this complication.

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, Sketch of the situation 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, $$\boxed{t = \frac{L}{v\cos\alpha} + \frac{s}{u}.}$$

I hope with these more detailed solution hints, you can answer the concrete questions you posed yourself (for example, you should now be able to tell that the system the published solution confusingly calls "moving" is the rest frame of the goal).

  • $\begingroup$ Thanks! I had forgotten that I had posted this problem, but your explanation is very clear and truly helpful. $\endgroup$
    – TheUnknown
    Jul 29, 2023 at 21:54

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