The equation given for the path of the ball is:

$y = -0.1x^2 + 2x + 1$

where $y$ represents the height of the ball in meters and $x$ represents the horizontal distance traveled in meters.

The goal is to determine if the ball will hit the ceiling, which is 10.6 meters high. To do this, we need to find the maximum height the ball reaches during its flight.

Step 1: Finding the maximum height

Since the equation for the ball's path is a quadratic function of the form $y = ax^2 + bx + c$, the maximum or minimum point of a quadratic equation occurs at the vertex. The x-coordinate of the vertex is given by the formula:

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

For the given equation $y = -0.1x^2 + 2x + 1$, the values of $a$ and $b$ are:

• $a = -0.1$
• $b = 2$

Substituting these values into the formula:

$x = -\frac{2}{2(-0.1)} = \frac{2}{0.2} = 10$

So, the horizontal distance where the maximum height occurs is at $x = 10$ meters.

Step 2: Calculating the maximum height

To find the maximum height, substitute $x = 10$ back into the original equation:

$y = -0.1(10)^2 + 2(10) + 1$ $y = -0.1(100) + 20 + 1$ $y = -10 + 20 + 1$ $y = 11 \text{ meters}$

Step 3: Conclusion

The maximum height the ball reaches is 11 meters. Since the ceiling of the gymnasium is 10.6 meters, the ball will hit the roof because 11 meters exceeds the ceiling height.

Would you like further clarification or details on this? Here are 5 questions you could explore based on this problem:

1. How would the path change if the initial velocity were lower?
2. How is the vertex formula derived for a quadratic equation?
3. What is the significance of the coefficient of $x^2$ in the trajectory equation?
4. How can we calculate the time the ball stays in the air?
5. How would the answer change if the gymnasium ceiling were higher?

Tip: The vertex of a parabola represents the maximum or minimum value in many real-life scenarios, such as projectiles reaching their highest point.