Projectile motion in a match between Brazil and Argentina (8634)

, por F_y_Q

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In a friendly football match between Argentina and Brazil, with the score tied 1–1 in the 90th minute, the referee awards a free kick to Brazil at a distance of 32 m from the goal. The player taking the shot is capable of imparting a speed of 30 m/s to the ball, and the defensive wall formed by the Argentine players, with an average height of 1.80 m, is positioned 12 m away from the point of the kick.

Determine:

a) What should be the launch angle in order to place the ball in the upper left corner of the goal without the wall blocking the shot?

b) What should be the launch angle in order to place the ball in the lower left corner of the goal without the wall blocking the shot?

P.-S.

This is an oblique‑launch projectile problem, and the solution will be carried out step by step, explaining the necessary approximations.

a) Placing the ball in the upper left corner without the wall obstructing the shot.

First, you must determine the minimum launch angle required for the ball to pass above the defensive wall.

The equations of projectile motion are:

$$$ \left. \begin{aligned} &\color{forestgreen}{\bf x(t) = v_0\cdot t\cdot cos\ \theta} \\ &\color{forestgreen}{\bf y(t) = v_0\cdot t\cdot sen\ \theta - \dfrac{g}{2}\cdot t^2} \end{aligned} \right \}$$$

The ball will clear the wall if, at the horizontal distance where the wall is located, its height is greater than the average height of the players. The time it takes the ball to reach the wall is:

$$$ \text{t}_\text{b} = \dfrac{\text{d}_\text{b}}{\text{v}_0\cdot \text{cos}\ \theta}\ \to\ \color{forestgreen}{\bf t_b = \dfrac{12}{30\cdot cos\ \theta}}$$$

Substituting this expression for the time into the vertical‑position equation:

$$$ \text{y}(\text{t}_\text{b}) = \text{v}_0\left(\dfrac{12}{30\cdot \text{cos}\ \theta}\right)\cdot \text{sen}\ \theta - \dfrac{\text{g}}{2}\left(\dfrac{12}{30\cdot \text{cos}\ \theta}\right)^2\ \to\ \color{forestgreen}{\bf y(t_b) = 12\cdot tg\ \theta - 4.9\left(\dfrac{0.16}{cos^2\ \theta}\right)}$$$

To solve the previous equation, two very useful approximations for small angles may be applied:

$$$ \left. \begin{aligned} \text{tg}\ \theta &\approx \theta \\ \text{cos}\ \theta &\approx 1 \end{aligned} \right \} \ \to\ \color{forestgreen}{\bf y(t_b) = 12\theta - 4.9\cdot 0.16}$$$

Solving for the angle gives:

$$$ \theta \ge \dfrac{1.80 + 0.784}{12}\ \to\ \color{royalblue}{\bf \theta \ge 12.3^o}$$$

You now have the minimum angle required to clear the wall, but you must adjust the angle so that the ball reaches the goal at the height of the crossbar, which in a football goal is 2.44 m.

The time it takes the ball to reach the goal is:

$$$ \text{t}_\text{p} = \dfrac{\text{d}_\text{p}}{\text{v}_0\cdot \text{cos}\ \theta}\ \to\ \color{forestgreen}{\bf t_p = \dfrac{32}{30\cdot cos\theta}}$$$

Substituting into the vertical‑position equation, analogously to the previous case, yields:

$$$ \text{y}(\text{t}_\text{p}) = \text{v}_0 \left(\dfrac{32}{30\cdot \text{cos}\ \theta}\right)\cdot \text{sen}\ \theta - \dfrac{\text{g}}{2} \left(\dfrac{32}{30\cdot \text{cos}\ \theta} \right)^2\ \to\ \color{forestgreen}{\bf y(t_p) = 32\theta - 4.9\cdot 1.14}$$$

Solving for the angle:

$$$ \theta \ge \dfrac{2.44 + 5.59}{32} \quad \Rightarrow \quad \color{firebrick}{\boxed{\bf \theta \ge 14.4^o}}$$$



b) Placing the ball in the lower left corner without the wall obstructing the shot.

The condition for clearing the wall is the same as in the previous section, and the minimum angle required remains: $$$ \theta = 12.3^o$$$. The only change is the condition that the ball must satisfy upon reaching the goal: in this case, the height must be zero:

$$$ \text{y}(\text{t}_\text{p}) = 32\theta - 4.9\cdot 1.14 \quad \Rightarrow \quad \theta \ge \dfrac{5.59}{32} \quad \Rightarrow \quad \color{firebrick}{\boxed{\bf \theta \ge 10^o}}$$$



Observe that this condition is already included in the requirement for the ball to clear the wall, so the angle $$$ \theta = 12.3^o$$$ is sufficient.