The height $$ y $$ (in feet) of a ball thrown by a child is $$ y=-\frac{1}{16} x^{2}+4 x+5 $$ where $$ x $$ is the horizontal distance in feet from the point at which the ball is thrown. (a) How high is the ball when it leaves the child's hand? (b) What is the maximum height of the ball? (c) How far from the child does the ball strike the ground?

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To solve the problem, we will analyze the quadratic function given for the height of the ball:

$$
y = -\frac{1}{16} x^{2} + 4x + 5
$$

### Step 1: Extract Known Conditions
- The function is a quadratic equation in the form $$ y = ax^2 + bx + c $$ where:
  - $$ a = -\frac{1}{16} $$
  - $$ b = 4 $$
  - $$ c = 5 $$

### Part (a): How high is the ball when it leaves the child's hand?
To find the height when the ball leaves the child's hand, we evaluate $$ y $$ at $$ x = 0 $$:

$$
y(0) = -\frac{1}{16}(0)^{2} + 4(0) + 5
$$

Calculating this gives:

$$
y(0) = 5
$$

So, the height of the ball when it leaves the child's hand is **5 feet**.

### Part (b): What is the maximum height of the ball?
The maximum height of a quadratic function occurs at the vertex. The x-coordinate of the vertex can be found using the formula:

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

Substituting the values of $$ a $$ and $$ b $$:

$$
x = -\frac{4}{2 \cdot -\frac{1}{16}} = -\frac{4}{-\frac{1}{8}} = 32
$$

Now, we will substitute $$ x = 32 $$ back into the height equation to find the maximum height:

$$
y(32) = -\frac{1}{16}(32)^{2} + 4(32) + 5
$$

Calculating this step by step:

1. Calculate $$ (32)^{2} = 1024 $$
2. Calculate $$ -\frac{1}{16}(1024) = -64 $$
3. Calculate $$ 4(32) = 128 $$

Now substituting these values into the equation:

$$
y(32) = -64 + 128 + 5 = 69
$$

Thus, the maximum height of the ball is **69 feet**.

### Part (c): How far from the child does the ball strike the ground?
The ball strikes the ground when $$ y = 0 $$. We need to solve the equation:

$$
-\frac{1}{16} x^{2} + 4x + 5 = 0
$$

To solve this quadratic equation, we will use the quadratic formula:

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

Substituting $$ a = -\frac{1}{16} $$, $$ b = 4 $$, and $$ c = 5 $$:

1. Calculate the discriminant $$ b^{2} - 4ac $$:

$$
b^{2} = 4^{2} = 16
$$
$$
4ac = 4 \cdot -\frac{1}{16} \cdot 5 = -\frac{20}{16} = -\frac{5}{4}
$$
$$
b^{2} - 4ac = 16 - (-\frac{5}{4}) = 16 + \frac{5}{4} = \frac{64}{4} + \frac{5}{4} = \frac{69}{4}
$$

2. Now substitute into the quadratic formula:

$$
x = \frac{-4 \pm \sqrt{\frac{69}{4}}}{2 \cdot -\frac{1}{16}} = \frac{-4 \pm \frac{\sqrt{69}}{2}}{-\frac{1}{8}} = \frac{-4 \pm \frac{\sqrt{69}}{2}}{-\frac{1}{8}} = 32 \pm 4\sqrt{69}
$$

Since we are looking for the positive distance, we will take the positive root:

$$
x = 32 + 4\sqrt{69}
$$

Now, we will calculate $$ 4\sqrt{69} $$:

$$
\sqrt{69} \approx 8.3066 \implies 4\sqrt{69} \approx 33.2264
$$

Thus,

$$
x \approx 32 + 33.2264 \approx 65.2264
$$

Therefore, the ball strikes the ground approximately **65.23 feet** from the child.

### Summary of Answers:
(a) 5 feet  
(b) 69 feet  
(c) Approximately 65.23 feet
(a) The ball is 5 feet high when it leaves the child's hand. (b) The maximum height of the ball is 69 feet. (c) The ball strikes the ground approximately 65.23 feet from the child.

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