Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s.

Respuesta :

Answer:

No she cannot.

Explanation:

Let [tex]v_h[/tex] be the horizontal component of the ball velocity when it's kicked, assume no air resistance, this is a constant. Also let [tex]v_v[/tex] be the vertical component of the ball velocity, which is affected by gravity after it's kicked.

The time it takes to travel 95m accross the field is

[tex]t = 95 / v_h[/tex] or [tex]v_h = 95/t[/tex]

t is also the time it takes to travel up, and the fall down to the ground, which ultimately stops the motion. So the vertical displacement after time t is 0

[tex]s = v_vt + gt^2/2= 0[/tex]

where g = -9.8m/s2 in the opposite direction with [tex]v_v[/tex]

[tex]v_vt - 4.9t^2 = 0[/tex]

[tex]v_vt = 4.9t^2[/tex]

[tex]v_v = 4.9t[/tex]

Since the total velocity that the goal keeper can give the ball is 30m/s

[tex]v = v_v^2 + v_h^2 = 30^2 = 900[/tex]

[tex](4.9t)^2 + \left(\frac{95}{t})^2 = 900[/tex]

[tex]24.01t^2 + \frac{9025}{t^2} = 900[/tex]

Let substitute x = [tex]t^2[/tex] > 0

[tex]24.01 x + \frac{9025}{x} = 900[/tex]

We can multiply both sides by x

[tex]24.01 x^2 + 9025 = 900x[/tex]

[tex]24.01x^2 - 900x + 9025 = 0[/tex]

[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]t= \frac{900\pm \sqrt{(-900)^2 - 4*(24.01)*(9025)}}{2*(24.01)}[/tex]

As [tex](-900)^2 - 4*24.01*9025 = -56761 < 0[/tex]

The solution for this quadratic equation is indefinite

So it's not possible for the goal keeper to do this.