A cannonball fired out to sea from a shore battery follows a parabolic trajectory given by the graph of the equation
h(x) = 11x − 0.01x^2
where h(x) is the height of the cannonball above the water when it has traveled a horizontal distance of x feet. (a) What is the maximum height that the cannonball reaches?
How far does the cannonball travel horizontally before splashing into the water?

To solve this problem, we need to analyze the quadratic function given by the equation:

\[
h(x) = 11x - 0.01x^2
\]

### (a) Maximum Height of the Cannonball

The maximum height of the cannonball is found at the vertex of the parabola. For a quadratic function in the form \( h(x) = ax^2 + bx + c \), the x-coordinate of the vertex (where the maximum or minimum value occurs) is given by:

\[
x = -\frac{b}{2a}
\]

Here, the coefficients are:
- \( a = -0.01 \)
- \( b = 11 \)

Substituting these values into the formula:

\[
x = -\frac{11}{2(-0.01)} = \frac{11}{0.02} = 550 \text{ feet}
\]

So, the maximum height occurs when the cannonball has traveled 550 feet horizontally.

To find the maximum height, substitute \( x = 550 \) back into the equation \( h(x) \):

\[
h(550) = 11(550) - 0.01(550)^2
\]

Calculate this:

\[
h(550) = 6050 - 0.01(302500) = 6050 - 3025 = 3025 \text{ feet}
\]

Thus, the maximum height the cannonball reaches is \( \boxed{3025 \text{ feet}} \).

### (b) Horizontal Distance Before Splashing Into the Water

The cannonball splashes into the water when its height \( h(x) \) is 0. Set \( h(x) = 0 \) and solve for \( x \):

\[
11x - 0.01x^2 = 0
\]

Factor the equation:

\[
x(11 - 0.01x) = 0
\]

This gives two solutions:

1. \( x = 0 \) (which corresponds to the initial firing position)
2. \( 11 - 0.01x = 0 \)

Solve for \( x \):

\[
0.01x = 11 \implies x = \frac{11}{0.01} = 1100 \text{ feet}
\]

So, the cannonball travels \( \boxed{1100 \text{ feet}} \) horizontally before splashing into the water.

Would you like more details or have any questions?

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Here are 8 related questions you might find useful:

1. How do you find the time it takes for the cannonball to reach its maximum height?
2. What is the initial velocity of the cannonball in the vertical direction?
3. How would the maximum height change if the coefficient of \(x^2\) were different?
4. What is the equation of the trajectory if the cannonball were fired from a height of 10 feet above the water?
5. How do you determine the range of a projectile if fired at an angle other than horizontal?
6. What is the significance of the vertex of a parabola in real-world problems like this?
7. Can you find the height of the cannonball at any given horizontal distance?
8. What are the implications of changing the angle of fire on the horizontal distance traveled?

**Tip:** The vertex of a parabola gives crucial information about the maximum or minimum point, which is often useful in physical problems involving projectiles.

Learn how to determine the maximum height and horizontal distance traveled by a cannonball fired in projectile motion. Solve quadratic equations and use vertex formulas with step-by-step explanations. Suitable for high school students.

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