To find the maximum height of the ball, we need to determine the vertex of the parabola given by the height function:

$h(T) = \frac{2}{7} (T + 3)(7 - T)$

First, we can rewrite the height function in standard quadratic form $h(T) = aT^2 + bT + c$ by expanding it:

$h(T) = \frac{2}{7} (T + 3)(7 - T)$

Expand the product inside the parentheses:

$h(T) = \frac{2}{7} (7T - T^2 + 21 - 3T)$ $h(T) = \frac{2}{7} (-T^2 + 4T + 21)$

Distribute $\frac{2}{7}$:

$h(T) = \frac{2}{7}(-T^2) + \frac{2}{7}(4T) + \frac{2}{7}(21)$ $h(T) = -\frac{2}{7} T^2 + \frac{8}{7} T + 6$

Now, the height function is in standard quadratic form:

$h(T) = -\frac{2}{7} T^2 + \frac{8}{7} T + 6$

The maximum height of a parabola $y = ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$. In this case, $a = -\frac{2}{7}$ and $b = \frac{8}{7}$:

$T = -\frac{b}{2a}$ $T = -\frac{\frac{8}{7}}{2 \cdot -\frac{2}{7}}$ $T = -\frac{\frac{8}{7}}{-\frac{4}{7}}$ $T = \frac{8}{7} \cdot \frac{7}{4}$ $T = 2$

Now we can find the maximum height by substituting $T = 2$ back into the height function:

$h(2) = \frac{2}{7} (2 + 3)(7 - 2)$ $h(2) = \frac{2}{7} \cdot 5 \cdot 5$ $h(2) = \frac{2}{7} \cdot 25$ $h(2) = \frac{50}{7}$ $h(2) \approx 7.14$

Therefore, the maximum height of the ball is approximately 7.14 units.

Would you like more details on this solution or have any other questions?

Here are 5 more questions you might ask:

1. How do you find the time of flight of the ball?
2. What is the initial velocity of the ball if kicked from the pier?
3. How do you find the derivative of the height function to determine the maximum height?
4. What is the significance of the vertex of a parabola in this context?
5. How do you find the total distance traveled by the ball?

Tip: When dealing with quadratic equations, finding the vertex is crucial for understanding the maximum or minimum values. Use the vertex formula $T = -\frac{b}{2a}$ to find the time at which the maximum or minimum height occurs.