## Statement

difficulty Determine the angle in relation to the horizontal, that you must kick a ball to the goal so that it scores barely touching the upper goal post, which is located at a height of 2.45 m, and 9 m away from the starting point. The ball is launched with a velocity of 82 km/h. Notice that the ball must be at the highest point of its trajectory to enter near the top corner of the goal.

## Solution

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:
• The initial velocity vector can be written as:

${\stackrel{\to }{v}}_{0}={v}_{0x}·\stackrel{\to }{i}+{v}_{0y}·\stackrel{\to }{j}={v}_{0}·\mathrm{cos}\left(\alpha \right)·\stackrel{\to }{i}+{v}_{0}·\mathrm{sin}\left(\alpha \right)·\stackrel{\to }{j}$

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

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

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

Resolution

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

$\stackrel{\to }{r}\left(t\right)=\left({v}_{0x}·t\right)·\stackrel{\to }{i}+\left({v}_{0y}·t-\frac{1}{2}·g·{t}^{2}\right)·\stackrel{\to }{j}=\left({v}_{0}·\mathrm{cos}\left(\alpha \right)·t\right)·\stackrel{\to }{i}+\left({v}_{0}·\mathrm{sin}\left(\alpha \right)·t-\frac{1}{2}·g·{t}^{2}\right)·\stackrel{\to }{j}$

The velocity vector is given by the expression:

$\stackrel{\to }{v}\left(t\right)={v}_{0x}·\stackrel{\to }{i}+\left({v}_{0y}-g·t\right)·\stackrel{\to }{j}={v}_{0}·\mathrm{cos}\left(\alpha \right)·\stackrel{\to }{i}+\left({v}_{0}·\mathrm{sin}\left(\alpha \right)-g·t\right)·\stackrel{\to }{j}$

When the ball enters the goal, it satisfies that:

${\stackrel{\to }{r}}_{f}=9·\stackrel{\to }{i}+2.45·\stackrel{\to }{j} ; {\stackrel{\to }{v}}_{f}={v}_{0x}·\stackrel{\to }{i}+0·\stackrel{\to }{j}$

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

$\left\{\begin{array}{l}0={v}_{0}·\mathrm{sin}\left(\alpha \right)-g·t\\ 9={v}_{0}·\mathrm{cos}\left(\alpha \right)·t\end{array}\right\\left\{\begin{array}{l}0={v}_{0}·\mathrm{sin}\left(\alpha \right)-g·t\\ t=\frac{9}{{v}_{0}·\mathrm{cos}\left(\alpha \right)}\end{array}\right\\left\{\begin{array}{l}0={v}_{0}·\mathrm{sin}\left(\alpha \right)-g·\frac{9}{{v}_{0}·\mathrm{cos}\left(\alpha \right)}\\ t=\frac{9}{{v}_{0}·\mathrm{cos}\left(\alpha \right)}\end{array}\right\⇒\phantom{\rule{0ex}{0ex}}⇒g·\frac{9}{{v}_{0}·\mathrm{cos}\left(\alpha \right)}={v}_{0}·\mathrm{sin}\left(\alpha \right)⇒88.2={22.78}^{2}·\mathrm{sin}\left(\alpha \right)·\mathrm{cos}\left(\alpha \right)$

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

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

Solving we get:

## Formulas worksheet

These are the main formulas that you must know to solve this exercise. If you are not clear about their meaning, we recommend you to check the theory in the corresponding sections. Furthermore, you will find in them, under the Formulas tab, the codes that will allow you to integrate these equations in external programs like Word or Mathematica.

Formulas
Related sections
$x={v}_{x}\cdot t={v}_{0}·\mathrm{cos}\left(\alpha \right)·t$
$y=\mathrm{H}+{\mathrm{v}}_{0\mathrm{y}}·\mathrm{t}-\frac{1}{2}·g·{t}^{2}=\mathrm{H}+{\mathrm{v}}_{0}·\mathrm{sin}\left(\mathrm{\alpha }\right)·\mathrm{t}-\frac{1}{2}·g·{t}^{2}$
${v}_{x}={v}_{0x}={v}_{0}·\mathrm{cos}\left(\alpha \right)$
${v}_{y}={v}_{0y}-g\cdot t={v}_{0}·\mathrm{sin}\left(\alpha \right)-g\cdot t$