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.