Keith plays hurling - Leaving Cert Mathematics - Question 10 - 2022

Substituting t = 2 seconds into the equation: $h(2) = -2(2)^2 + 5(2) + 1 = -8 + 10 + 1 = 3 ext{ m}.$
Therefore, the ball was 3 metres high when it was caught.

To find when the ball reached a height of 3 metres, set the height function equal to 3: $3 = -2t^2 + 5t + 1.$
Rearranging gives: $0 = -2t^2 + 5t - 2.$
Using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4(-2)(-2)}}{2(-2)} = \frac{-5 \pm \sqrt{25 - 16}}{-4} = \frac{-5 \pm 3}{-4}.$
So:

1. $t = \frac{-2}{-4} = 0.5 ext{ seconds}$
2. $t = \frac{-8}{-4} = 2 ext{ seconds}$
   Therefore, the ball passed over the halfway line at 0.5 seconds.

To find the critical points, take the derivative of the height function: $\frac{dh}{dt} = -4t + 5.$
Setting this equal to zero to find when the height is at its maximum: $0 = -4t + 5 \\ => t = 1.25 ext{ seconds}.$ Hence, it took the ball 1.25 seconds to reach its greatest height.

The coordinates of the points based on the described quadratic function are:

1. Initial point: (0, 1)
2. Maximum height point: (2, 5)
3. Ground point: (4, 0)

The graph of the function $y = k(t)$ can be sketched as a smooth curve passing through these points, touching the x-axis at t = 4 seconds.