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The problem you're working on involves the trajectory of a projectile launched from a cliff. The height of the projectile \( h(x) \) as a function of the horizontal distance \( x \) from the cliff is given by the equation:

\[
h(x) = -\frac{32x^2}{60^2} + x + 220
\]

This represents the vertical motion of the projectile, and the goal is to answer several questions based on this function.

Let's address each part of the problem:

### (c) At what horizontal distance will the projectile strike the water?

You are asked to find when the projectile strikes the water. This happens when \( h(x) = 0 \), so you need to solve the quadratic equation:

\[
-\frac{32x^2}{60^2} + x + 220 = 0
\]

From the image, it seems this value has already been calculated as \( x = 226 \) feet.

### (d) Graphing the function

You are asked to choose the correct graph that represents the function \( h(x) \) over the interval \( 0 \leq x \leq 230 \) with a graphing window set as:
- \( x \) range: [0, 230]
- \( y \) range: [0, 270]

From the options provided, it seems option **B** is selected, and it appears correct. The graph starts at a height of 220 feet (the initial height of the cliff), rises slightly, and then curves downward, reaching the water.

### (e) When the height of the projectile is 100 feet above the water, how far is it from the cliff?

This part asks you to find \( x \) when \( h(x) = 100 \). Plugging this into the equation, you get:

\[
100 = -\frac{32x^2}{60^2} + x + 220
\]

Simplifying:

\[
- \frac{32x^2}{3600} + x + 220 = 100
\]
\[
- \frac{32x^2}{3600} + x + 120 = 0
\]

Now, solve this quadratic equation to find \( x \), which represents the horizontal distance from the cliff when the projectile is 100 feet above the water.

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Would you like me to walk you through solving this quadratic equation step-by-step? 

Here are five related questions to expand on this:
1. How does the angle of launch affect the trajectory?
2. How would air resistance change the equation of motion?
3. What is the maximum height the projectile reaches?
4. How can you calculate the time it takes for the projectile to hit the water?
5. What does the discriminant of the quadratic equation tell you about the projectile’s path?

**Tip**: Always double-check your graph window settings when using a graphing calculator to ensure you're seeing the correct portion of the function.

A projectile is fired from a cliff 220 feet above the water at an inclination of 45° to the horizontal, with a muzzle velocity of 60 feet per second. The height h(x) of the projectile above the water is given by the equation h(x) = (-32x^2 / 60^2) + x + 220. Use this information to answer the following questions.

Solve a projectile motion problem involving a quadratic function. The equation h(x) = (-32x^2 / 60^2) + x + 220 models the height of a projectile above water. Learn to solve for the distance it strikes the water and how far it is from the cliff when it reaches 100 feet above the water.

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