Elevation angle in a parabolic kick

Data

• The initial velocity v₀ = 82 km/h = 22.78 m/s
• Height of the goal post (final height) y = 2.45 m
• Horizontal distance between the starting point and the goal (final distance) x = 9 m

Preliminary considerations

• It is a parabolic motion. The parabolic motion is a composition of two motions:
  • uniform rectilinear motion in the x-axis
  • uniformly accelerated rectilinear motion in the y-axis

• The initial velocity vector can be written as:

  $${\overrightarrow{v}}_0=v_{0x}·\overrightarrow{i}+v_{0y}·\overrightarrow{j}=v_0·\cos \left(\alpha \right)·\overrightarrow{i}+v_0·\sin \left(\alpha \right)·\overrightarrow{j}$$

  $\alpha$  which is precisely the angle we are looking for

• The ball enters the goal at the highest point of the trajectory, that is, when v_y = 0

• Consider the value of gravity g = 9.8 m/s²

Resolution

The equation of position in parabolic motion is given by the expression

$$\overrightarrow{r}\left(t\right)=\left(v_{0x}·t\right)·\overrightarrow{i}+\left(v_{0y}·t-\frac{1}{2}·g·t^2\right)·\overrightarrow{j}=\left(v_0·\cos \left(\alpha \right)·t\right)·\overrightarrow{i}+\left(v_0·\sin \left(\alpha \right)·t-\frac{1}{2}·g·t^2\right)·\overrightarrow{j}$$ 

The velocity vector is given by the expression:

$$\overrightarrow{v}\left(t\right)=v_{0x}·\overrightarrow{i}+(v_{0y}-g·t)·\overrightarrow{j}=v_0·\cos \left(\alpha \right)·\overrightarrow{i}+(v_0·\sin \left(\alpha \right)-g·t)·\overrightarrow{j}$$ 

When the ball enters the goal, it satisfies that:

$${\overrightarrow{r}}_f=9·\overrightarrow{i}+2.45·\overrightarrow{j}\hspace{0.17em};\hspace{0.17em}{\overrightarrow{v}}_f=v_{0x}·\overrightarrow{i}+0·\overrightarrow{j}$$

Equating the expressions of the equation of position and of velocity to the previous vectors, we can use the y component of the velocity v_y and the x component of the position vector r_x to determine the launching angle:

$$\begin{gathered} \left\{\begin{array}{l}0=v_0·\sin \left(\alpha \right)-g·t\\ 9=v_0·\cos \left(\alpha \right)·t\end{array}\right.\left\{\begin{array}{l}0=v_0·\sin \left(\alpha \right)-g·t\\ t=\frac{9}{v_0·\cos \left(\alpha \right)}\end{array}\right.\left\{\begin{array}{l}0=v_0·\sin \left(\alpha \right)-g·\frac{9}{v_0·\cos \left(\alpha \right)}\\ t=\frac{9}{v_0·\cos \left(\alpha \right)}\end{array}\right.\Rightarrow \\ \Rightarrow g·\frac{9}{v_0·\cos \left(\alpha \right)}=v_0·\sin \left(\alpha \right)\Rightarrow 88.2={22.78}^2·\sin \left(\alpha \right)·\cos \left(\alpha \right) \end{gathered}$$

Now we need to remember the following trigonometric identity to solve the equation:

$$\sin \left(\alpha \right)·\cos \left(\alpha \right)=\frac{\sin \left(2\alpha \right)}{2}$$ 

Solving we get:

$$88.2={22.78}^2·\sin \left(\alpha \right)·\cos \left(\alpha \right)\Rightarrow 0.169=\frac{\sin \left(2\alpha \right)}{2}\Rightarrow 2\alpha ={\sin }^{-1}\left(0.339\right)\Rightarrow \boxed{\alpha =0.17 rad}$$