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To solve for the maximum height reached by the ball, we need to analyze the given equation for height, which is:

\[
h(t) = -2t^2 + 12t + 2.3
\]

This is a quadratic equation of the form:

\[
h(t) = at^2 + bt + c
\]

where \(a = -2\), \(b = 12\), and \(c = 2.3\).

### Step 1: Find the time \(t\) at which the ball reaches its maximum height
In a quadratic equation, the maximum (or minimum) value occurs at \( t = \frac{-b}{2a} \). Using this formula:

\[
t = \frac{-12}{2(-2)} = \frac{-12}{-4} = 3 \, \text{seconds}
\]

### Step 2: Substitute \(t = 3\) into the height equation to find the maximum height
Now, we substitute \(t = 3\) into the equation \(h(t)\):

\[
h(3) = -2(3)^2 + 12(3) + 2.3
\]
\[
h(3) = -2(9) + 36 + 2.3
\]
\[
h(3) = -18 + 36 + 2.3
\]
\[
h(3) = 18 + 2.3 = 20.3 \, \text{meters}
\]

### Conclusion:
The maximum height reached by the ball is **20.3 meters**, to the nearest tenth.

Would you like further clarification on any of the steps?

Here are 5 related questions:
1. How do we know that this quadratic equation has a maximum and not a minimum?
2. What happens to the ball's height after it reaches its maximum?
3. How can we find when the ball hits the ground using this equation?
4. How would the height equation change if the initial velocity were different?
5. How do air resistance and gravity influence the motion described by this equation?

**Tip:** In quadratic functions, if the coefficient of \(t^2\) is negative, the function has a maximum value; if it's positive, it has a minimum.

A ball is thrown vertically upward with an initial velocity of 12 m/s. Its height, h(t) metres after t seconds, is given by the equation h(t) = -2t^2 + 12t + 2.3. The maximum height, to the nearest tenth, reached by the ball is ______ m.

Learn how to find the maximum height reached by a ball thrown vertically upward with an initial velocity of 12 m/s. Using the quadratic equation h(t) = -2t^2 + 12t + 2.3, we determine the vertex, which gives us the maximum height. Suitable for students in Grades 9-12.

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