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.              When the height of the projectile is 100 feet above the​ water, how far is it from the​ cliff?
xequals

enter your response here feet

We are given the height function of the projectile above the water:

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

where \( h(x) \) represents the height of the projectile above the water and \( x \) represents the horizontal distance from the cliff.

We are asked to find \( x \) when \( h(x) = 100 \) feet. This means we need to solve the equation:

\[
100 = \frac{-32x^2}{55^2} + x + 210
\]

Let's solve this equation step by step.

First, rearrange the equation:

\[
100 = \frac{-32x^2}{3025} + x + 210
\]

Subtract 210 from both sides:

\[
100 - 210 = \frac{-32x^2}{3025} + x
\]

\[
-110 = \frac{-32x^2}{3025} + x
\]

Now, let's solve this equation for \( x \).The solutions for \( x \) are:

\[
x = \frac{3025}{64} - \frac{55 \sqrt{17105}}{64} \quad \text{or} \quad x = \frac{3025}{64} + \frac{55 \sqrt{17105}}{64}
\]

Let's calculate these approximate values to determine the horizontal distance from the cliff when the projectile is 100 feet above the water.The two possible solutions for \( x \) are approximately:

\[
x \approx -65.13 \, \text{feet} \quad \text{or} \quad x \approx 159.66 \, \text{feet}
\]

Since a negative distance from the cliff doesn't make sense in this context, the projectile is approximately **159.66 feet** from the cliff when it is 100 feet above the water.

Would you like further clarification or details on any part of this solution?

Here are 5 follow-up questions:
1. How would you calculate the maximum height of the projectile?
2. At what horizontal distance will the projectile hit the water?
3. How does the inclination angle affect the height equation?
4. What would change if the muzzle velocity were doubled?
5. How long does the projectile remain in the air before it reaches the water?

**Tip:** When solving quadratic equations, always consider the physical context to determine which solutions are meaningful.

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. When the height of the projectile is 100 feet above the​ water, how far is it from the​ cliff?

Learn how to solve a projectile motion equation where a projectile is fired from a 210-foot cliff at 45° and a velocity of 55 ft/s. Find the horizontal distance when the height is 100 feet. Suitable for Grades 10-12, this problem covers key algebra concepts, projectile motion, and solving quadratic equations.

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