The given equation for the height of the ball is:

$h(t) = 3 + 12t - 2t^2$

This is a quadratic function, and the maximum height occurs at the vertex of the parabola (since the coefficient of $t^2$ is negative, the parabola opens downwards).

For a quadratic function of the form $h(t) = at^2 + bt + c$, the time at which the maximum height occurs is given by:

$t_{\text{max}} = \frac{-b}{2a}$

In this case, $a = -2$, $b = 12$, and $c = 3$. Using the formula:

$t_{\text{max}} = \frac{-12}{2(-2)} = \frac{-12}{-4} = 3 \text{ seconds}$

Now, substitute $t = 3$ into the original equation to find the maximum height:

$h(3) = 3 + 12(3) - 2(3)^2$ $h(3) = 3 + 36 - 2(9)$ $h(3) = 3 + 36 - 18 = 21 \text{ metres}$

Thus, the maximum height of the ball is 21 metres.

Would you like more details or have any questions?

Here are 5 related questions:

1. What is the significance of the vertex in a quadratic equation?
2. How does changing the value of $a$ in $at^2 + bt + c$ affect the shape of the parabola?
3. What is the initial height of the ball at $t = 0$?
4. How can you determine when the ball hits the ground?
5. What other physical interpretations can be drawn from this quadratic function?

Tip: The vertex form of a quadratic equation can make it easier to identify the maximum or minimum value directly.