Let's solve this problem step-by-step based on the given quadratic equation:

The height $y$ of the ball is described by: $y = -\frac{1}{2}x^2 + 2x + 4$ where $x$ is the horizontal distance in feet.

1. How high was the ball when it left the child’s hand?

To find the height when the ball left the child's hand, we need to evaluate $y$ at $x = 0$, which is the initial position.

$y = -\frac{1}{2}(0)^2 + 2(0) + 4 = 4$

Answer: The ball was 4 feet high when it left the child’s hand.

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2. How high was the ball when it was at its maximum height?

The maximum height of a quadratic function $y = ax^2 + bx + c$ occurs at the vertex, where $x = -\frac{b}{2a}$.

In this equation:

• $a = -\frac{1}{2}$
• $b = 2$

So, $x = -\frac{2}{2 \times -\frac{1}{2}} = -\frac{2}{-1} = 2$

Now, substitute $x = 2$ back into the equation to find the height at this point: $y = -\frac{1}{2}(2)^2 + 2(2) + 4$ $y = -\frac{1}{2}(4) + 4 + 4$ $y = -2 + 4 + 4 = 6$

Answer: The ball reached a maximum height of 6 feet.

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3. How far from the child did the ball strike the ground?

To find the distance from the child where the ball struck the ground, we need to find the value of $x$ when $y = 0$.

Set $y = 0$: $0 = -\frac{1}{2}x^2 + 2x + 4$

To make this equation easier to solve, multiply through by -2: $0 = x^2 - 4x - 8$

Now, we can solve this quadratic equation using the quadratic formula: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1}$ $x = \frac{4 \pm \sqrt{16 + 32}}{2}$ $x = \frac{4 \pm \sqrt{48}}{2}$ $x = \frac{4 \pm 4\sqrt{3}}{2}$ $x = 2 \pm 2\sqrt{3}$

This gives two solutions, but we only consider the positive one since $x$ represents a distance: $x = 2 + 2\sqrt{3}$

Approximating $\sqrt{3} \approx 1.732$: $x \approx 2 + 2 \times 1.732 = 2 + 3.464 = 5.464$

Answer: The ball struck the ground approximately 5.46 feet from the child.

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Summary

1. Initial height: 4 feet
2. Maximum height: 6 feet
3. Distance from the child where the ball struck the ground: Approximately 5.46 feet

Would you like more details on any of these steps, or do you have additional questions?

Here are some related questions for further exploration:

1. How would the height change if the initial throw was stronger, modifying the equation?
2. What would happen to the maximum height if the child threw the ball at a steeper angle?
3. How would you find the time it takes for the ball to reach its maximum height?
4. What is the ball's vertical velocity when it leaves the child's hand?
5. Can we calculate the total time the ball is in the air based on this equation?

Tip: To determine the highest point of a parabola, remember to use the vertex formula $x = -\frac{b}{2a}$ for quadratic equations.