A Ball Is Thrown Vertically Upward. After $t$ Seconds, Its Height $h$ (in Feet) Is Given By The Function $h(t)=44t-16t^2$. What Is The Maximum Height That The Ball Will Reach?Do Not Round Your Answer.Height:

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Introduction

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When a ball is thrown vertically upward, its height can be modeled using a quadratic function. In this case, the height of the ball after $t$ seconds is given by the function $h(t) = 44t - 16t^2$. The goal is to find the maximum height that the ball will reach. To do this, we need to analyze the function and determine its maximum value.

Understanding the Function

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The given function is a quadratic function in the form of $h(t) = at^2 + bt + c$, where $a = -16$, $b = 44$, and $c = 0$. The graph of a quadratic function is a parabola that opens upward or downward. In this case, since $a = -16$ is negative, the parabola opens downward, indicating that the function has a maximum value.

Finding the Maximum Value

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To find the maximum value of the function, we need to find the vertex of the parabola. The vertex of a parabola in the form of $h(t) = at^2 + bt + c$ is given by the formula $t = -\frac{b}{2a}$. Plugging in the values of $a$ and $b$, we get:

$t = -\frac{44}{2(-16)}$ $t = -\frac{44}{-32}$ $t = \frac{11}{8}$

Calculating the Maximum Height

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Now that we have found the value of $t$ at the vertex, we can plug it back into the original function to find the maximum height:

$h(t) = 44t - 16t^2$ $h(\frac{11}{8}) = 44(\frac{11}{8}) - 16(\frac{11}{8})^2$ $h(\frac{11}{8}) = 44(\frac{11}{8}) - 16(\frac{121}{64})$ $h(\frac{11}{8}) = \frac{484}{8} - \frac{1936}{64}$ $h(\frac{11}{8}) = \frac{484}{8} - \frac{121}{16}$ $h(\frac{11}{8}) = \frac{484 \times 2}{16} - \frac{121}{16}$ $h(\frac{11}{8}) = \frac{968}{16} - \frac{121}{16}$ $h(\frac{11}{8}) = \frac{847}{16}$

Conclusion

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The maximum height that the ball will reach is $\frac{847}{16}$ feet.

Discussion

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The maximum height of the ball is determined by the vertex of the parabola. The vertex is the point on the parabola where the function changes from increasing to decreasing or vice versa. In this case, the vertex is at $t = \frac{11}{8}$ seconds, and the maximum height is $\frac{847}{16}$ feet.

Final Answer

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The final answer is $\boxed{\frac{847}{16}}$.

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Introduction

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In our previous article, we explored the problem of a ball thrown vertically upward and modeled its height using a quadratic function. We found that the maximum height the ball will reach is $\frac{847}{16}$ feet. In this article, we will answer some common questions related to this problem.

Q&A

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Q: What is the initial velocity of the ball?

A: The initial velocity of the ball is not given in the problem. However, we can find it by looking at the coefficient of $t$ in the function $h(t) = 44t - 16t^2$. The coefficient of $t$ represents the initial velocity, which is $44$ feet per second.

Q: What is the acceleration due to gravity?

A: The acceleration due to gravity is represented by the coefficient of $t^2$ in the function $h(t) = 44t - 16t^2$. The coefficient of $t^2$ is $-16$, which represents the acceleration due to gravity. Since the acceleration due to gravity is $-16$, it means that the ball is accelerating downward at a rate of $16$ feet per second squared.

Q: How long does it take for the ball to reach its maximum height?

A: To find the time it takes for the ball to reach its maximum height, we need to find the value of $t$ at the vertex of the parabola. We already found that the vertex is at $t = \frac{11}{8}$ seconds.

Q: What is the maximum height the ball will reach if it is thrown from a height of $h_0$ feet?

A: If the ball is thrown from a height of $h_0$ feet, the new function will be $h(t) = h_0 + 44t - 16t^2$. To find the maximum height, we need to find the vertex of the new parabola. The vertex will be at $t = -\frac{b}{2a}$, where $a = -16$ and $b = 44 + \frac{h_0}{t}$. Plugging in the values, we get:

$t = -\frac{44 + \frac{h_0}{t}}{2(-16)}$ $t = -\frac{44 + \frac{h_0}{t}}{-32}$ $t = \frac{44 + \frac{h_0}{t}}{32}$

Solving for $t$, we get:

$t = \frac{32(44 + \frac{h_0}{t})}{32}$ $t = 44 + \frac{h_0}{t}$ $t^2 - 44t - h_0 = 0$

Using the quadratic formula, we get:

$t = \frac{-(-44) \pm \sqrt{(-44)^2 - 4(1)(-h_0)}}{2(1)}$ $t = \frac{44 \pm \sqrt{1936 + 4h_0}}{2}$

The maximum height will be at $t = \frac{44 + \sqrt{1936 + 4h_0}}{2}$ seconds.

Q: What is the maximum height the ball will reach if it is thrown with an initial velocity of $v_0$ feet per second?

A: If the ball is thrown with an initial velocity of $v_0$ feet per second, the new function will be $h(t) = v_0t - 16t^2$. To find the maximum height, we need to find the vertex of the new parabola. The vertex will be at $t = -\frac{b}{2a}$, where $a = -16$ and $b = v_0$. Plugging in the values, we get:

$t = -\frac{v_0}{2(-16)}$ $t = \frac{v_0}{32}$

The maximum height will be at $t = \frac{v_0}{32}$ seconds.

Conclusion

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In this article, we answered some common questions related to the problem of a ball thrown vertically upward. We found the initial velocity, acceleration due to gravity, time it takes for the ball to reach its maximum height, and the maximum height the ball will reach if it is thrown from a height of $h_0$ feet or with an initial velocity of $v_0$ feet per second.

Final Answer

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The final answers are:

• Initial velocity: $44$ feet per second
• Acceleration due to gravity: $-16$ feet per second squared
• Time it takes for the ball to reach its maximum height: $\frac{11}{8}$ seconds
• Maximum height the ball will reach if it is thrown from a height of $h_0$ feet: $\frac{847}{16} + h_0$
• Maximum height the ball will reach if it is thrown with an initial velocity of $v_0$ feet per second: $\frac{v_0^2}{64}$