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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?

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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.

Ms. Runnalls hits a massive pop-up. The height h, in metres, of the ball is given by h = 3 + 12t - 2t², where t is in seconds. What is the maximum height of the ball?

Learn how to find the maximum height of a ball hit by Ms. Runnalls using the quadratic function h = 3 + 12t - 2t². This example demonstrates how to apply the vertex formula to find the peak height. Suitable for students in Grades 9-11.

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