To find the maximum height the ball will reach, we need to determine the vertex of the parabola represented by the equation

$y = -\frac{x^2}{100} + x + 100$

This is a quadratic equation in the form:

$y = ax^2 + bx + c$

where $a = -\frac{1}{100}$, $b = 1$, and $c = 100$. In a quadratic function $y = ax^2 + bx + c$, the x-coordinate of the vertex, which gives the maximum height in this case, is given by:

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

Step 1: Find the x-coordinate of the vertex

Plugging in the values of $a$ and $b$:

$x = -\frac{1}{2 \cdot \left(-\frac{1}{100}\right)} = \frac{1}{\frac{2}{100}} = 50$

So, the maximum height occurs when $x = 50$ feet from the building.

Step 2: Substitute $x = 50$ into the equation to find $y$

Now, substitute $x = 50$ back into the equation to find the corresponding $y$-value:

$y = -\frac{(50)^2}{100} + 1 \cdot 50 + 100$

Calculating each term:

1. $-\frac{(50)^2}{100} = -\frac{2500}{100} = -25$
2. $1 \cdot 50 = 50$
3. $+100$

Adding these together:

$y = -25 + 50 + 100 = 125$

Conclusion

The maximum height the ball will reach is 125 feet.

---

Would you like additional details or have any questions?

Here are 5 related questions to expand on this problem:

1. How do we interpret the path of the ball using the equation given?
2. What would change if the ball was thrown with a different initial velocity?
3. How would the maximum height be affected if the building were taller or shorter?
4. What is the significance of the vertex in a projectile motion problem?
5. How would we calculate the total distance traveled by the ball horizontally?

Tip: In projectile motion problems, the vertex represents the highest point the object will reach.