Question

A football is kicked off the ground. After traveling a horizontal distance of 24m, it just passes over a tree that is 3m tall before hitting the ground 28m from where it was kicked. a) Consider the ground to be the x-axis and assume the vertex lies on the y-axis. Determine the equation of the quadratic equation that models the path of the ball. b)Determine the maximum height of the ball. c)Suppose the football is kicked from a starting point at the origin. Develop a new equation of the quadratic function of the quadratic function that represents the football.

Answer 1

Note: The explanation of problems b) and c) are attached as images because of the maximum limit of 5,000 characters.

Answers:

a)
`y = -(1)/(32) x^2 + (49)/(8)`
b)
`(49)/(8) \text{ meters}`
c)
`y = -(1)/(32) (x-14)^2 + (49)/(8)`

Explanations:

a) Note that the path where the ball travels when kicked off the ground is the parabola.
Let (h,k) be the vertex of the parabola. Since the vertex is in y-axis, the x-coordinate of the vertex is 0. So, h = 0.
The equation of the parabola with vertex (h, k) is given by

`y = a(x - h)^2 + k` (1) Since h = 0, we can replace this to equation (1) so that

`y = a(x - h)^2 + k\\ y = a(x - 0)^2 + k\\ y = ax^2 + k` (2)

To solve for a and k, we need to get the coordinates of the points of parabola. In the problem, it is possible to determine the coordinates of the points where the ball is kicked off from the ground and where the ball ended hitting the ground and the point where the ball hits the tree which are points of the parabola.
Since the ground is in x-axis and the ball is kicked off from the ground and ended hitting the ground, the parabola intersects the x-axis at the points where the ball is kicked off from the ground and where the ball ended hitting the ground.
Let

A = point where the ball is kicked off from the ground
B = point where the ball ended hitting the ground
M be the midpoint of A and B.

Note that the x-coordinate of the vertex of the parabola is the same as the x-coordinate of M. Since the x-coordinate of the vertex of the parabola is 0, the x-coordinate of M is 0. Moreover, since the ground is in x-axis, A and B are in the x-axis and so M is in x-axis. So, the y-coordinate of M is 0 because the y-coordinate of any points of x-axis is 0. Thus, the coordinates of M are (0,0).
As mentioned earlier, the y-coordinate of any points in x-axis is 0. Since A and B both lie in x-axis, their y-coordinate is 0. Now, we let j be the x-coordinate of A and k be the x-coordinate of B so that (j, 0) are the coordinates of A and (k, 0) are the coordinates of B.

Since the ball traveled with a horizontal distance of 28 meters when it is kicked off until it hit the ground, the horizontal distance from A to B is 28. Since the coordinates of A and B are (j, 0) and (k, 0) respectively, using the distance formula the distance from A and B is given by


`d = √((j-k)^2 + (0 - 0)^2) \\ 28 = √((j-k)^2) \\ (j-k)^2 = 28^2 \\ \boxed{j - k = \pm 28}`

Since A is the point where the ball is kicked off and B is the point where the ball hits the ground, A is in the left of B. Since A and B are both in x-axis and j and k are the x-coordinates of A and B respectively, j < k. Thus, j - k is negative and so

j - k = -28 (3)

Because M is the midpoint of A and B, the x-coordinate of M is
`(j+k)/(2)`. Since the x-coordinate of M is 0,


`(j+k)/(2) = 0` (4)

Multiplying both sides of equation (4) by 2, we have

j + k = 0 (5)

Now, we add equations (3) and (5) so that

2j = -28
j = -14

Substituting the value of j in (5), we have

j + k = 0
-14 + k = 0
k = 14

Hence the coordinates of A are (-14,0) and the coordinates of B are (14,0). Since point A is in parabola, we can substitute its coordinates of A in equation (2) with x = -14 and y = 0 so that


`y = ax^2 + k \\ 0 = a(-14)^2 + k \\ 0 = 196a + k \\ \boxed{196a + k =0}` (6)

Note that, if we substitute the coordinate of B, we also arrive at equation (6), which is useless.

So, we need to determine the coordinates of another point of parabola, which is the point where the ball hits the tree. We call this point C. Since the tree is 3 meters high and the ground is in x-axis, the y-coordinate of point C is 3. Let c be the x-coordinate of C. Since the ball hits the tree after travelling a horizontal distance of 24 meters, the horizontal distance from A to C is 24. Moreover, the horizontal distance is the difference of x-coordinates of two points. Since C is the right of A and the x-coordinate of A is -14

c - (-14) = 24
c + 14 = 24
c = 10

Hence, the coordinates of C are (10, 3). Substituting this to equation (2), we have


`y = ax^2 + k \\ 3 = a(10)^2 + k \\ 3 = 100a + k \\ \boxed{100a + k =3}` (7)

Subtracting equation (6) to (7), we have


`(196a + k)-(100a + k) = 0-3 \\ 96a = -3 \\ \boxed{a = -(1)/(32) }`

Substituting the value of a in equation (6) we have


`196a + k = 0 \\ 196(-(1)/(32)) + k = 0 \\ \\ -(49)/(8) + k =0 \\ \\ \boxed{k = (49)/(8) }`

Using equation (2) and the values of a and k, the equation of the parabola is given by


`\boxed{y = -(1)/(32) x^2 + (49)/(8) }`