A Rocket Is Launched From A Tower. The Height Of The Rocket, \[$ Y \$\] In Feet, Is Related To The Time After Launch, \[$ X \$\] In Seconds, By The Given Equation:$\[ Y = -16x^2 + 194x + 149 \\]Using This Equation, Find The

Feb 28, 2025 by ADMIN 224 views

A Rocket is Launched from a Tower: A Mathematical Exploration

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Introduction

In this article, we will delve into the world of mathematics and explore the trajectory of a rocket launched from a tower. The height of the rocket, denoted by ${$ y $}$, is related to the time after launch, denoted by ${$ x $}$, by the given equation: ${ y = -16x^2 + 194x + 149 }$ . Our goal is to use this equation to find the maximum height reached by the rocket and the time at which it occurs.

Understanding the Equation

The given equation is a quadratic equation in the form of ${ y = ax^2 + bx + c }$ , where ${ a = -16 }$ , ${ b = 194 }$ , and ${ c = 149 }$ . This type of equation represents a parabola, which is a U-shaped curve. The vertex of the parabola represents the maximum or minimum point of the curve.

Finding the Vertex

To find the vertex of the parabola, we can use the formula: ${ x = -\frac{b}{2a} }$ . Plugging in the values of ${ a }$ and ${ b }$ , we get: ${ x = -\frac{194}{2(-16)} }$ . Simplifying this expression, we get: ${ x = -\frac{194}{-32} }$ , which further simplifies to: ${ x = \frac{194}{32} }$ . Therefore, the time at which the rocket reaches its maximum height is ${ x = \frac{194}{32} }$ seconds.

Finding the Maximum Height

Now that we have found the time at which the rocket reaches its maximum height, we can plug this value back into the original equation to find the maximum height reached by the rocket. Substituting ${ x = \frac{194}{32} }$ into the equation, we get: ${ y = -16\left(\frac{194}{32}\right)^2 + 194\left(\frac{194}{32}\right) + 149 }$ . Simplifying this expression, we get: ${ y = -16\left(\frac{194^2}{32^2}\right) + 194\left(\frac{194}{32}\right) + 149 }$ . Further simplifying, we get: ${ y = -16\left(\frac{37636}{1024}\right) + 194\left(\frac{194}{32}\right) + 149 }$ . This simplifies to: ${ y = -\frac{601856}{1024} + \frac{37636}{16} + 149 }$ . Converting the fractions to have a common denominator, we get: ${ y = -\frac{601856}{1024} + \frac{37636 \times 64}{16 \times 64} + \frac{149 \times 1024}{1024} }$ . This simplifies to: ${ y = -\frac{601856}{1024} + \frac{2408832}{1024} + \frac{152576}{1024} }$ . Combining the fractions, we get: ${ y = \frac{2408832 + 152576 - 601856}{1024} }$ . This simplifies to: ${ y = \frac{1951152}{1024} }$ . Finally, simplifying this fraction, we get: ${ y = 1896 }$ .

Conclusion

In conclusion, the maximum height reached by the rocket is 1896 feet, and it occurs at a time of ${ x = \frac{194}{32} }$ seconds. This problem demonstrates the use of quadratic equations to model real-world situations and find maximum or minimum values.

Step-by-Step Solution

Step 1: Understand the Equation

The given equation is a quadratic equation in the form of ${ y = ax^2 + bx + c }$ , where ${ a = -16 }$ , ${ b = 194 }$ , and ${ c = 149 }$ .

Step 2: Find the Vertex

Step 3: Find the Maximum Height

Step 4: Simplify the Expression

Simplifying this expression, we get: ${ y = -16\left(\frac{194^2}{32^2}\right) + 194\left(\frac{194}{32}\right) + 149 }$ . Further simplifying, we get: ${ y = -16\left(\frac{37636}{1024}\right) + 194\left(\frac{194}{32}\right) + 149 }$ . This simplifies to: ${ y = -\frac{601856}{1024} + \frac{37636}{16} + 149 }$ .

Step 5: Combine the Fractions

Converting the fractions to have a common denominator, we get: ${ y = -\frac{601856}{1024} + \frac{37636 \times 64}{16 \times 64} + \frac{149 \times 1024}{1024} }$ . This simplifies to: ${ y = -\frac{601856}{1024} + \frac{2408832}{1024} + \frac{152576}{1024} }$ .

Step 6: Simplify the Expression

Combining the fractions, we get: ${ y = \frac{2408832 + 152576 - 601856}{1024} }$ . This simplifies to: ${ y = \frac{1951152}{1024} }$ .

Step 7: Final Simplification

Finally, simplifying this fraction, we get: ${ y = 1896 }$ .

Final Answer

The maximum height reached by the rocket is 1896 feet, and it occurs at a time of ${ x = \frac{194}{32} }$ seconds.

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Introduction

In our previous article, we explored the trajectory of a rocket launched from a tower using a quadratic equation. We found that the maximum height reached by the rocket is 1896 feet, and it occurs at a time of ${ x = \frac{194}{32} }$ seconds. In this article, we will answer some frequently asked questions related to this problem.

Q&A

Q: What is the equation of the parabola that represents the trajectory of the rocket?

A: The equation of the parabola is ${ y = -16x^2 + 194x + 149 }$ .

Q: How do you find the vertex of the parabola?

A: To find the vertex of the parabola, we can use the formula: ${ x = -\frac{b}{2a} }$ . Plugging in the values of ${ a }$ and ${ b }$ , we get: ${ x = -\frac{194}{2(-16)} }$ . Simplifying this expression, we get: ${ x = -\frac{194}{-32} }$ , which further simplifies to: ${ x = \frac{194}{32} }$ .

Q: What is the maximum height reached by the rocket?

A: The maximum height reached by the rocket is 1896 feet.

Q: At what time does the rocket reach its maximum height?

A: The rocket reaches its maximum height at a time of ${ x = \frac{194}{32} }$ seconds.

Q: How do you find the maximum height reached by the rocket?

A: To find the maximum height reached by the rocket, we can plug the value of ${ x }$ back into the original equation. Substituting ${ x = \frac{194}{32} }$ into the equation, we get: ${ y = -16\left(\frac{194}{32}\right)^2 + 194\left(\frac{194}{32}\right) + 149 }$ .

Q: What is the significance of the vertex of the parabola?

A: The vertex of the parabola represents the maximum or minimum point of the curve. In this case, the vertex represents the maximum height reached by the rocket.

Q: Can you explain the concept of quadratic equations in simple terms?

A: A quadratic equation is a type of equation that represents a parabola. The parabola is a U-shaped curve that can be either upward-facing or downward-facing. The vertex of the parabola represents the maximum or minimum point of the curve.