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Part 1
A projectile is fired from a cliff 210 feet above the water at an inclination of​ 45° to the​ horizontal, with a muzzle velocity of 55 feet per second. The height h of the projectile above the water is given by
h left parenthesis x right parenthesis equals StartFraction negative 32 x squared Over left parenthesis 55 right parenthesis squared EndFraction plus x plus 210
where x is the horizontal distance of the projectile from the face of the cliff. Use this information to answer the following.
Question content area bottom
Part 1
​(a) At what horizontal distance from the face of the cliff is the height of the projectile a​ maximum?
xequals

enter your response here feet
​(Round to the nearest foot as​ needed.)

We are given the height function for the projectile above the water as follows:

\[
h(x) = -\frac{32x^2}{55^2} + x + 210
\]

where:
- \( x \) is the horizontal distance of the projectile from the face of the cliff,
- \( h(x) \) is the height of the projectile above the water.

To find the horizontal distance \( x \) at which the height is maximized, we need to find the vertex of this quadratic function. For a quadratic equation in the form:

\[
h(x) = ax^2 + bx + c
\]

the horizontal distance \( x \) at which the height is maximum can be found using the formula for the vertex:

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

### Step 1: Identify the coefficients
The given equation is:

\[
h(x) = -\frac{32}{55^2}x^2 + x + 210
\]

So the coefficients are:
- \( a = -\frac{32}{55^2} \)
- \( b = 1 \)

### Step 2: Apply the vertex formula

Now, let's plug in the values into the vertex formula:

\[
x = -\frac{1}{2 \times -\frac{32}{55^2}}
\]

We'll now compute this step by step.The horizontal distance at which the height of the projectile is maximized is approximately 47 feet (rounded to the nearest foot).

Would you like more details or have any further questions?

Here are 5 related questions for deeper understanding:
1. How do you determine whether the vertex of a quadratic function gives a maximum or a minimum value?
2. What is the maximum height reached by the projectile at this horizontal distance?
3. How does the muzzle velocity affect the trajectory of the projectile?
4. How would the height function change if the cliff were taller or shorter?
5. How can you derive the time of flight from the given height function?

**Tip:** The sign of the quadratic term \(a\) in a parabola determines whether the function opens upward (minimum) or downward (maximum).

A projectile is fired from a cliff 210 feet above the water at an inclination of​ 45° to the​ horizontal, with a muzzle velocity of 55 feet per second. The height h of the projectile above the water is given by h(x) = -(32x^2)/(55^2) + x + 210 where x is the horizontal distance of the projectile from the face of the cliff. At what horizontal distance from the face of the cliff is the height of the projectile a maximum?

Solve a projectile motion problem involving a projectile fired from a 210-foot cliff with an inclination of 45° and a muzzle velocity of 55 feet per second. Learn how to find the maximum height using quadratic equations and the vertex formula. Suitable for Grades 10-12.

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