## Statement

difficulty

Minute 90 of the game... Lopera approaches the ball to make a direct free kick 40 meters from the goal, takes two steps back and kiiicks. The ball takes off at an elevation of 20°... and GOOOOALLL!!! GOOOOOOOALLL!!!! The ball goes in through the top corner at a height of 1.70 m!!!. After hearing this radio broadcast, can you answer the following questions?

a) From Lopera’s kick to scoring the goal, how long did it take? and what was the initial velocity of the ball at the moment of the kick?
b) What is the maximum height reached by the ball?
c) How fast was the ball going when it reached the goal?

## Solution

Question a)

At the moment when the ball reaches the goal, x = 40 m and y = 1.7 m. Substituting in the equations of position for parabolic motion:

Question b)

When the y component of the velocity (vy) is 0 then it means that the ball will be at the highest point of the parable. Remember that as the ball ascends its velocity in the y-axis decreases until it becomes zero and it start increasing in the opposite direction as it descends

Now we are ready, applying the equation of position in the y-axis, and substituting the time we just calculated, we determine the maximum height reached:

Question c)

Knowing that the ball reached the goal in 1.61 s, we get that is velocity was:

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