Question

A soccer ball is kicked from the ground by the goalie and lands 40 meters from where it waskicked. The ball's path is that of a parabola, and the ball reaches a maximum height of 3 meters,How far from where the ball was kicked will it have a height of 2 meters? (Round your answer tothe nearest tenth.)

Answer 1

Given:

The distance travelled by the ball is, d = 40 m.

The maximum height the ball reached is, h = 3 m.

The objective is to find the distance travelled by the ball at a height of h' = 2 m.

. (vertex)

n, the

The value of x can be calculated using the general form of the equation of parabola,

`y=a(x-p)^2+q`

Here, (p,q) stands for the vertex of the parabola (20,3).

To find the value of a, consider a coordinate (0,0) and substitute the obtained values in the general equation of parabola.

`\begin{gathered} 0=a(0-20)^2+3 \\ 0=a(400)+3 \\ -400a=3 \\ a=-(3)/(400) \end{gathered}`

Now, consider the coordinate of the required distance x of the ball, (x,2).

Substitute the above coordinate, the value of a and the vertex in the equation of parabola.

`\begin{gathered} y=a(x-p)^2+q \\ 2=-(3)/(400)(x-20)^2+3^{} \\ 2=-(3)/(400)(x-20)^2+3 \\ 2-3=-(3)/(400)(x-20)^2 \\ -1=-(3)/(400)(x-20)^2 \\ ((400)/(3))=(x-20)^2 \end{gathered}`

To solve the square on RHS, take square root on both sides of the equation.

`\begin{gathered} \sqrt[]{(400)/(3)}=\sqrt[]{(x-20)^2} \\ 11.5=(x-20) \\ x=11.5+20 \\ x=31.5\text{ m} \end{gathered}`

Since, the height of 2 m of ball can be obtained at either side of the vertex.

So the distance between the vertex and the required position is

`\begin{gathered} c=31.5-20 \\ c=11.5 \end{gathered}`

Then, the other possible position of the ball is,

`\begin{gathered} x^(\prime)=20-c \\ x^(\prime)=20-11.5 \\ x^(\prime)=8.5 \end{gathered}`

Hence, the height of the ball will be 2m, either at a distance of 8.5m from origin or 31.5m from the origin.