The given equation is:

$y = -\frac{1}{12}x^2 + 6x + 5$

where $y$ represents the height of the ball in feet, and $x$ represents the horizontal distance in feet from the point of throw. We'll solve each part step-by-step.

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(a) How high is the ball when it leaves the child's hand?

When the ball leaves the child’s hand, the horizontal distance $x = 0$.

Substitute $x = 0$ into the equation:

$y = -\frac{1}{12}(0)^2 + 6(0) + 5 = 5$

Answer: The ball is 5 feet high when it leaves the child’s hand.

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(b) What is the maximum height of the ball?

The height is a quadratic function, and the maximum height occurs at the vertex of the parabola. For a quadratic equation in the form:

$y = ax^2 + bx + c$

the $x$-coordinate of the vertex is given by:

$x = -\frac{b}{2a}$

Here:

• $a = -\frac{1}{12}$
• $b = 6$

$x = -\frac{6}{2 \cdot -\frac{1}{12}} = -\frac{6}{-\frac{1}{6}} = 36 \, \text{feet (horizontal distance at maximum height)}.$

Substitute $x = 36$ into the equation to find $y$:

$y = -\frac{1}{12}(36)^2 + 6(36) + 5$

$y = -\frac{1}{12}(1296) + 216 + 5$

$y = -108 + 216 + 5 = 113$

Answer: The maximum height of the ball is 113 feet.

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(c) How far from the child does the ball strike the ground?

The ball strikes the ground when $y = 0$. Solve the quadratic equation:

$0 = -\frac{1}{12}x^2 + 6x + 5$

Multiply through by $-12$ to eliminate fractions:

$0 = x^2 - 72x - 60$

Solve this using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here:

• $a = 1$
• $b = -72$
• $c = -60$

$x = \frac{-(-72) \pm \sqrt{(-72)^2 - 4(1)(-60)}}{2(1)}$

$x = \frac{72 \pm \sqrt{5184 + 240}}{2}$

$x = \frac{72 \pm \sqrt{5424}}{2}$

$x = \frac{72 \pm 73.65}{2}$

Two solutions: $x = \frac{72 + 73.65}{2} = 72.83 \quad \text{(positive root, valid for distance)}$ $x = \frac{72 - 73.65}{2} = -0.83 \quad \text{(negative root, not valid)}$

Answer: The ball strikes the ground 72.83 feet from the child.

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Final Answers:

(a) 5 feet
(b) 113 feet
(c) 72.83 feet

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Do you want detailed calculations or visual graphs for better understanding?
Here are 5 related questions to explore:

1. How do you determine the axis of symmetry for a parabola?
2. How would the equation change if the ball was thrown from a higher point?
3. What are the effects of changing the coefficient of $x^2$ on the trajectory?
4. How can we derive the time it takes for the ball to reach its maximum height?
5. How does the quadratic formula apply to solving real-world problems?

Tip: Always double-check units in word problems—they’re key to accurate solutions!