To find the maximum height of the ball, we need to determine the vertex of the quadratic function:

$y = -\frac{1}{2}x^2 + 4x + 5$

In a quadratic equation of the form $y = ax^2 + bx + c$, the $x$-coordinate of the vertex, which represents the point where the function reaches its maximum (or minimum, depending on the sign of $a$), is given by:

$x = -\frac{b}{2a}$

In this case:

• $a = -\frac{1}{2}$
• $b = 4$

Substitute these values:

$x = -\frac{4}{2 \cdot -\frac{1}{2}} = -\frac{4}{-1} = 4$

Now, substitute $x = 4$ back into the equation to find $y$:

$y = -\frac{1}{2}(4)^2 + 4 \cdot 4 + 5$ $y = -\frac{1}{2} \cdot 16 + 16 + 5$ $y = -8 + 16 + 5$ $y = 13$

Therefore, the maximum height of the ball is 13 feet.

Would you like more details or have any questions?

Here are some related questions:

1. How would you find the time it takes for the ball to reach this maximum height?
2. What does the negative coefficient of $x^2$ signify in terms of the ball's trajectory?
3. How would you find the time when the ball hits the ground?
4. How can you determine the initial height of the ball when it was thrown?
5. How would the equation change if the ball were thrown with a different initial speed?

Tip: The maximum or minimum of a parabola occurs at its vertex, so for any quadratic function, finding the vertex can give you the peak value if the parabola opens downwards.