The Walla Walla Little League teams use a baseball throwing machine to help train 10-year old players to catch high pop-ups.  Assume the quadratic function 
h
(
t
)
=
−
16
t
2
+
48
t
+
3.5
h(t)=−16t 
2
 +48t+3.5  can be used to model the height of the ball in feet, 
t
t seconds after it is thrown.


The given quadratic function to model the height of the ball after it is thrown by a machine is:

\[
h(t) = -16t^2 + 48t + 3.5
\]

where:
- \( h(t) \) represents the height of the ball (in feet) at time \( t \) (in seconds),
- \( t \) is the time in seconds after the ball is thrown.

### 1. Maximum Height
The ball's height is modeled by a quadratic function, and since the leading coefficient (−16) is negative, the parabola opens downward, indicating that there is a maximum point, which corresponds to the highest point the ball will reach.

The time \( t_{\text{max}} \) at which the ball reaches its maximum height can be found using the vertex formula:

\[
t_{\text{max}} = \frac{-b}{2a}
\]

In this case:
- \( a = -16 \),
- \( b = 48 \).

Substitute these values into the formula:

\[
t_{\text{max}} = \frac{-48}{2(-16)} = \frac{-48}{-32} = 1.5 \, \text{seconds}
\]

Now, substitute \( t = 1.5 \) back into the height equation to find the maximum height:

\[
h(1.5) = -16(1.5)^2 + 48(1.5) + 3.5
\]
\[
h(1.5) = -16(2.25) + 72 + 3.5
\]
\[
h(1.5) = -36 + 72 + 3.5
\]
\[
h(1.5) = 39.5 \, \text{feet}
\]

### 2. Time When the Ball Hits the Ground
To find when the ball hits the ground, we need to solve for \( t \) when \( h(t) = 0 \):

\[
-16t^2 + 48t + 3.5 = 0
\]

This is a quadratic equation, and we can solve it using the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where:
- \( a = -16 \),
- \( b = 48 \),
- \( c = 3.5 \).

Substitute these values into the formula:

\[
t = \frac{-48 \pm \sqrt{48^2 - 4(-16)(3.5)}}{2(-16)}
\]
\[
t = \frac{-48 \pm \sqrt{2304 + 224}}{-32}
\]
\[
t = \frac{-48 \pm \sqrt{2528}}{-32}
\]
\[
t = \frac{-48 \pm 50.28}{-32}
\]

Thus, we get two possible values for \( t \):

\[
t_1 = \frac{-48 + 50.28}{-32} = \frac{2.28}{-32} \approx -0.071 \, \text{(ignore, as time cannot be negative)}
\]
\[
t_2 = \frac{-48 - 50.28}{-32} = \frac{-98.28}{-32} \approx 3.07 \, \text{seconds}
\]

So, the ball hits the ground at \( t \approx 3.07 \) seconds.

### Summary:
- The ball reaches its maximum height of **39.5 feet** at \( t = 1.5 \) seconds.
- The ball hits the ground at \( t \approx 3.07 \) seconds.

Would you like more details on any part of this solution? 

Here are 5 related questions:
1. How would the trajectory change if the initial velocity was increased?
2. How does the coefficient of \( t^2 \) affect the motion of the ball?
3. Can you find the velocity of the ball at any given time using the function?
4. How would air resistance alter the equation of motion?
5. How can you model the time when the ball reaches half of its maximum height?

**Tip**: The vertex of a parabola in a quadratic function provides the maximum or minimum value, which is crucial in motion problems like this.

The Walla Walla Little League teams use a baseball throwing machine to help train 10-year-old players to catch high pop-ups. Assume the quadratic function h(t) = -16t^2 + 48t + 3.5 can be used to model the height of the ball in feet, t seconds after it is thrown.

Explore how to use a quadratic function to model the height of a baseball thrown by a machine. Learn to find the maximum height and when the ball hits the ground using key algebraic formulas. This problem is perfect for students in Grades 8-10 and covers projectile motion concepts.

Math Problem Statement

The Walla Walla Little League teams use a baseball throwing machine to help train 10-year old players to catch high pop-ups. Assume the quadratic function h ( t )

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