The height 𝑦 (in feet) of a ball thrown by a child is
𝑦=-116𝑥2+2𝑥+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?    feet

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

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

Let's address each part of the problem one by one.

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

The height of the ball when it leaves the child's hand corresponds to the value of \( y \) when \( x = 0 \).

Given the equation:
\[ y = -116x^2 + 2x + 5 \]

Substitute \( x = 0 \):
\[ y = -116(0)^2 + 2(0) + 5 \]
\[ y = 5 \]

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

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

To find the maximum height, we need to determine the vertex of the quadratic equation \( y = -116x^2 + 2x + 5 \).

The x-coordinate of the vertex is given by:
\[ x = -\frac{b}{2a} \]

Here, \( a = -116 \) and \( b = 2 \):
\[ x = -\frac{2}{2(-116)} \]
\[ x = -\frac{2}{-232} \]
\[ x = \frac{1}{116} \]

Now, substitute \( x = \frac{1}{116} \) back into the equation to find the y-coordinate:
\[ y = -116\left(\frac{1}{116}\right)^2 + 2\left(\frac{1}{116}\right) + 5 \]
\[ y = -116\left(\frac{1}{13456}\right) + \frac{2}{116} + 5 \]
\[ y = -\frac{116}{13456} + \frac{2}{116} + 5 \]
\[ y = -\frac{1}{116} + \frac{1}{58} + 5 \]
\[ y = -\frac{1}{116} + \frac{2}{116} + 5 \]
\[ y = \frac{1}{116} + 5 \]
\[ y \approx 5.0086 \]

So, the maximum height of the ball is approximately \( 5.009 \) feet.

### (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 quadratic equation for \( y = 0 \):

\[ -116x^2 + 2x + 5 = 0 \]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = -116, \, b = 2, \, c = 5 \]

\[ x = \frac{-2 \pm \sqrt{2^2 - 4(-116)(5)}}{2(-116)} \]
\[ x = \frac{-2 \pm \sqrt{4 + 2320}}{-232} \]
\[ x = \frac{-2 \pm \sqrt{2324}}{-232} \]
\[ x = \frac{-2 \pm 48.21}{-232} \]

This gives two solutions:
\[ x_1 = \frac{-2 + 48.21}{-232} \approx -0.198 \]
\[ x_2 = \frac{-2 - 48.21}{-232} \approx 0.217 \]

The negative solution doesn't make sense in this context, so we take the positive one:
\[ x \approx 0.217 \]

So, the ball strikes the ground approximately \( 0.217 \) feet from the child.

Would you like any more details or have any questions?

Here are 5 questions you can ask:
1. Can you verify the solution for the maximum height calculation?
2. How do you solve quadratic equations using the quadratic formula?
3. What are the steps to derive the quadratic formula?
4. Can you explain the physical significance of each term in a quadratic equation related to projectile motion?
5. How can we check if a quadratic equation will have real, repeated, or complex roots?

**Tip:** When solving quadratic equations, double-check your calculations for potential arithmetic errors to ensure accuracy.

Explore how to solve a quadratic equation describing the height of a ball in projectile motion. Find out the initial height, maximum height, and landing distance using detailed step-by-step calculations. Suitable for high school students studying quadratic equations and projectile motion.

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