Soccer trick shot

Classical Mechanics Level 3

Billy exercises shots on a goal wall to improve his soccer skills. The goal wall has two holes that serve as targets for the ball with diameter $d = 22\, \text{cm}$ . Since he already can hit both holes from a short distance relatively reliably, he is looking for a new challenge. Billy considers a trick shot in which he hits both holes with a single shot. He uses the fact that directly behind the goal wall is a brick wall, where the ball can bounce off. If he hits the hole at top left with a diagonal shot, then it is possible that the ball rebounds on the back wall and then reaches the second hole at bottom right. But what should the optimal shooting angle $\theta$ be? As a result, enter the numerical value for $\tan \theta$ (the initial slope of the trajectory) with an accuracy of two decimal places.

Details and Assumptions: Billy shoots at the goal from the white line, which is 6 meters apart from the goal line. Behind the goal wall is the brick wall at a distance of 1.5 meters. The two holes in the goal wall have a horizontal distance--with respect to their centers--of 1.6 meters, and a height of 1.35 and 0.35 meters above the ground, respectively. (All geometrical dimensions are also drawn in the diagram in units of centimeters.) The holes are sufficiently large to allow the ball to pass through. For the optimal shot, the ball should pass through the centers of the holes.

We assume that the ball does not rotate around its own axis and rebounds completely elastically on the back wall. The back wall is also completely flat and even (the brick pattern is in reality just a wallpaper). When the ball hits the back wall, it deforms to a hemisphere. We can therefore treat it as a mass point located at the center of the ball during this bounce. Although the ball must first be accelerated by Billy's foot to reach its initial speed, we assume that this acceleration phase is only negligibly short. Thus, the ball starts directly on the ground with a certain initial velocity and is subject only to the gravitational force (frictional forces are therefore not considered).

The answer is 0.50.

2 solutions

Markus Michelmann
Jan 8, 2018

The trajectory of the ball is simply a straight line connecting the points $A$ , $B$ and $C$ in plan view (horizontal plane). When the ball rebounds elastically at the brick wall at point $C$ , the normal component of its velocity is inverted ( $v_\perp' = - v_\perp$ ). Therefore, the angle of incidence $\phi$ is equal to the angle of reflection. Instead of explicitly considering the reflection of the ball, instead we mirror the track $\overline{CD}$ on the back wall and get the track $\overline{CD'}$ instead. We can therefore pretend, that the brick wall did not exist and instead the ball moves in a straight line on a second goal wall behind the first at point $D'$ .

We place the x-axis so that it is parallel to the trajectory of the ball. The points $A$ , $B$ and $D'$ correspond to the points $x = 0$ , $x = 4c$ and $x = 6c$ , respectively. The length $c$ is given by $c = \sqrt{150^2 + \left( \frac{160}{2}\right)^2 } \, \text{cm} = 170 \, \text{cm}$ Viewed from the side, the ball trajectory gives a parabola passing through the points $(x, z) = (0,11)$ , $(4c,135)$ and $(6c,35)$ in units of centimeters. (The initial height of 11 centimeters corresponds to the radius of the ball.) These three points set the trajectory clearly. So we describe the trajectory through the polynomial $h(x) = q \cdot \left(\frac{x}{c}\right)^2 + r \cdot \frac{x}{c} + s$ with an offset $s = 11\,\text{cm}$ , which we can read directly. For the constants $q$ and $r$ we get a system of linear equations: $\begin{aligned} 16 q + 4 r &= 124 \,\text{cm} & & (\text{point } B)\\ 36 q + 6 r &= 24 \,\text{cm} & & (\text{point } D') \end{aligned}$ which is solved by $q = - \frac{27}{2}\,\text{cm}, \quad r = 85 \,\text{cm}$ The slope $\tan \theta$ of the trajectory in point $A$ is given by the first derivative of the parabola: $\tan \theta = \left. \frac{d h}{dx} \right|_{x = 0} = \frac{r}{c} = \frac{1}{2}$

Does this solution not assume that the ball goes through the center of each hole? I think you can shave distance by allowing the ball close to the edges of the targets... Then "c" becomes ~166.68cm, and "r" becomes ~79.58cm, giving a value for TAN of 0.48 (i.e. a lower trajectory to account for the fact the ball can "hit" the targets diagonally closer than the centers).

Clearly the intent was your solution, but I'm curious if it's accurate (or if am I just incorrect).

Ryan Johnston - 3 years, 4 months ago

You're right, that the problem does not state clearly, that the trajectory is asked, which goes exactly through the middle of the two holes. The fact that the diameter $D$ of the holes is explicitly stated, although it plays no role in the solution, only adds to the confusion. (Originally, I wanted to ask if this trick shot is even possible, or if the ball hits the hole at such a shallow angle that he can not go through it.)

Markus Michelmann - 3 years, 4 months ago

I've edited the problem so that the shooter is aiming for an "optimal shot" in which the ball goes through the centers of the holes.

Andrew Hayes Staff - 3 years, 4 months ago

Why is the depth behind the goal 150? When the ball bounces, do we not assume that its center is still 11cm away from the wall? If you're going to include this offset for the starting point, it seems it should also be included there.

Lydia Nevin - 3 years, 4 months ago

I honestly did not think about that. (But now I pretend that it was meant to be, and quickly consider a justification.) If we assume, that the ball will be deformed into a hemisphere on impact with the wall, we would not need to consider the extra distance! Of course, if the ball was inflated extra heavily, so that it practically does not deform during the rebound, the entire ball radius must be added on the distance.

Of course, the reality is somewhere in between. But this point should be clarified in the problem.

Markus Michelmann - 3 years, 4 months ago

I've edited the problem statement to account for this. Thanks!

Andrew Hayes Staff - 3 years, 4 months ago

Wouldn't the ball go faster and longer after the balls air was compressed. Wouldn't it gain velocity? If it were to continue on with out it would loose momentum due to gravity? Wouldn't you want to have the loss of momentum factored in? Sorry I am not a super smart person.

Christina Kihn - 3 years, 4 months ago

I assumed that points B and D' have to obey equation of parabola in form y=11 + x tan(theta) - g/2 x^2/(Vo cos(theta))^2 where y is either 124 or 24 and x is either 680 (170 4) or 1020 (680 + 2 170). By eliminating term (Vo cos(theta))^2 results in linear equation for tangent theta.

Mirek Baudys - 3 years, 4 months ago

Very nice. I laughed at the compression to a hemisphere (quite big compression), but it made the point clear that it's basically a straight line to point D'. But then I started to solve it as a point mass and didn't start with y=11 at A.

A Former Brilliant Member - 3 years, 4 months ago

Mariano PerezdelaCruz
Jan 27, 2018

The core of the problem rest in realizing that the path of the ball is a parabola like were not existing the brick-wall. Then we take as belonging to the parabola the point that mirror of second hole with respect to the brick-wall. Therefore we have 3 points which define the parabola and its tangent at the kicking point.

Soccer trick shot

The answer is 0.50.

2 solutions

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