Question

A soccer player who is 27 feet from a goal attempted to kick the ball in into the goal. the flight of the ball is modeled by a parabola. the ball reached a maximum height of 10 feet when it was 15 feet from the soccer player. the goal has a height of 8 feet. will the soccer ball land in the goal?

Answer 1

The parabola is facing downward that is why it is going to take the equation,
 
                     x² = -4ay
If we take the point at which the maximum height is reached be the origin, the starting point of the parabola from the left is going to be (-15, -10). 
Substituting the values of x and y to the equation,
                (-15)² = -4(a)(-10)
                            a = 5.625

The goal is at the point equal to (27 - 15, 8 - 10) or (12, -2). We are to substitute these values to the equation and see if the equation is satisfied.
                   x² = -4(5.625)(y)
                   12² = -4(5.625)(-2)
                  144 ≠ 45

Thus, the ball will not land in the goal. 

Answer 2

Answer:

The ball will land inside the goal.

Step-by-step explanation:

Given : The flight of the ball is modeled by a parabola.

To find : Will the soccer ball land in the goal?

Solution :

Let the equation of the parabola be,

$y=a(x-h)^2+k$

Where, (h,k) are the vertex of the parabola,

According to question,

A soccer player who is 27 feet from a goal.

Let A be the point where player stand i.e, (27,0)

The height of the goal is 8 feet.

Let B be the point of height of goal i.e, (0,8)

The ball reached a maximum height of 10 feet when it was 15 feet from the soccer player.

The distance from goal to the distance 15 feet away from the player is

27-15=12 feet

Let C be the point with maximum height i.e, (12,10)

C is the vertex of the parabola and A is the point on the parabola.

Then, The equation of parabola is

$y=a(x-12)^2+10$

Point A satisfy the equation,

$0=a(27-12)^2+10$

$0=a(15)^2+10$

$a=-\frac{10}{225}$

Substitute back in equation,

$y=-\frac{10}{225}(x-12)^2+10$

Now, we find the y-intercept of the equation i.e, x=0,

$y=-\frac{10}{225}(0-12)^2+10$

$y=-\frac{10}{225}(144)+10$

$y=3.6$

If the y-intercept is greater than 8 then the ball will land outside the goal.

If the y-intercept is less than 8 then the ball will land inside the goal.

As 3.6<8

Therefore, The ball will land inside the goal.

Refer the attached figure below.