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To find the maximum height reached by the ball, we will analyze the given quadratic function \( h(t) = -2t^2 + 12t + 2.3 \), where \( h(t) \) is the height of the ball in meters and \( t \) is time in seconds.

### Steps:
1. **Identify the form of the quadratic equation**: 
   The equation is in the form \( h(t) = at^2 + bt + c \), where \( a = -2 \), \( b = 12 \), and \( c = 2.3 \).

2. **Find the time at which the maximum height occurs**:
   For a quadratic equation \( h(t) = at^2 + bt + c \), the time at which the maximum height occurs is given by the vertex formula:
   \[
   t_{\text{max}} = \frac{-b}{2a}
   \]
   Substituting \( a = -2 \) and \( b = 12 \):
   \[
   t_{\text{max}} = \frac{-12}{2(-2)} = \frac{-12}{-4} = 3 \, \text{seconds}.
   \]

3. **Calculate the maximum height**:
   Substitute \( t = 3 \) into the original equation to find the height at this time:
   \[
   h(3) = -2(3)^2

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 using the quadratic function h(t) = -2t^2 + 12t + 2.3. This problem covers key concepts such as quadratic functions and the vertex theorem, suitable for students in Grades 9-11.

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