5. Flight of soccer ball. AJbert kicks a soccer it reaches its from the ground. Two seconds later, it maximum height of 16 m and then lands on the ground after a further two seconds. (i) Given that the height, \( y \) metres, of the ball \( x \) seconds after it has been kicked can be modelled by the quadratic equation

We are told that AJbert kicks the soccer ball from the ground (so at x = 0 seconds the height is 0 m) and that after 2 seconds the ball reaches its maximum height of 16 m. It then lands back on the ground after a further 2 seconds—that is, at x = 4 seconds the height is again 0 m. Because the motion is modelled by a quadratic function of the form   y = ax² + bx + c we can use the given conditions to determine the coefficients. Step 1. Initial condition (kick from the ground): At x = 0, y = 0. Substituting:   0 = a(0)² + b(0) + c ⟹ c = 0. Step 2. Landing condition (ball hits the ground at x = 4 seconds): At x = 4, y = 0. With c = 0 this gives:   0 = a(4)² + b(4)   0 = 16a + 4b ⟹ Divide both sides by 4:   0 = 4a + b ⟹ b = –4a. Step 3. Vertex condition (maximum height at x = 2 seconds): Because the vertex of a quadratic a x² + bx occurs at x = –b/(2a), and we are told the vertex is at x = 2, we have:   2 = –b/(2a) Substitute b = –4a:   2 = –(–4a)/(2a) = 4a/(2a) = 2. This verifies our relation between a and b. We also know at x = 2 the height y = 16. Substitute x = 2 into y = ax² + bx:   16 = a(2)² + b(2) = 4a + 2b. Replace b with –4a:   16 = 4a + 2(–4a) = 4a – 8a = –4a, which gives:   –4a = 16 ⟹ a = –4. Then b = –4a = –4(–4) = 16. Step 4. Write the quadratic function: Substituting a, b, and c into y = ax² + bx + c, we obtain:   y = –4x² + 16x. Alternate form (vertex form): You can also express it as   y = –4(x – 2)² + 16, which makes the vertex at (2, 16) clear. Thus, the height y (in meters) of the ball at time x (in seconds) is given by:   y = –4x² + 16x.