The function f(t) = –16t2 + 250 represents the distance of a dropped object from the ground at any time, t. The equation g(t) = 15t represents the height of the object if it is tossed upward at a rate of 15 feet per second. The sum of the functions can be used to model the distance the dropped object is from the ground if it is tossed upward at a rate of 15 feet per second before it begins its descent.

To solve the problem and understand the behavior of the object's motion when it is tossed upward and then falls back to the ground, let's explore the sum of the functions \( f(t) \) and \( g(t) \).

### Given Functions:
1. **Distance of a dropped object from the ground**:  
   \[
   f(t) = -16t^2 + 250
   \]
   Here, \( t \) represents time in seconds, and \( f(t) \) gives the height of the object at time \( t \) as it falls under gravity (assuming initial height of 250 feet).

2. **Height of the object if tossed upward at 15 feet per second**:  
   \[
   g(t) = 15t
   \]
   This represents the height gained by the object due to the initial upward velocity.

### Sum of the Functions:
The sum of the functions \( h(t) = f(t) + g(t) \) models the height of the object from the ground if it is initially tossed upward before it begins its descent.

\[
h(t) = f(t) + g(t) = (-16t^2 + 250) + 15t
\]
\[
h(t) = -16t^2 + 15t + 250
\]

### Interpretation:
The function \( h(t) = -16t^2 + 15t + 250 \) represents the height of the object at any time \( t \) after being tossed upward. 

- **\( -16t^2 \)** accounts for the effect of gravity, pulling the object back down.
- **\( 15t \)** accounts for the initial upward velocity.
- **\( 250 \)** is the initial height from which the object is tossed.

### Analyzing the Motion:
To better understand the object's motion, we can analyze the function \( h(t) \):
- **Maximum height**: This occurs at the vertex of the parabola described by \( h(t) \).
- **Time to reach maximum height**: The vertex of a parabola \( at^2 + bt + c \) occurs at \( t = -\frac{b}{2a} \).

Let's find the vertex:
\[
t = -\frac{15}{2(-16)} = \frac{15}{32} \approx 0.46875 \text{ seconds}
\]

At this time, the object reaches its maximum height.

- **Maximum height**: Substitute \( t = \frac{15}{32} \) into \( h(t) \) to find the maximum height.

\[
h\left(\frac{15}{32}\right) = -16\left(\frac{15}{32}\right)^2 + 15\left(\frac{15}{32}\right) + 250
\]
\[
\text{Calculate to find the exact height.}
\]

### Conclusion:
The function \( h(t) = -16t^2 + 15t + 250 \) describes the height of the object at any time \( t \). The object is tossed upward, reaches a maximum height, and then falls back down to the ground.

Would you like to see a detailed calculation of the maximum height, or do you have any other specific questions?

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### Related Questions:
1. What is the object's initial velocity and how does it affect its motion?
2. How long does it take for the object to hit the ground after being tossed upward?
3. What is the significance of the coefficient \( -16 \) in the function?
4. How would the motion change if the object was tossed upward with a different initial velocity?
5. Can we calculate the total time the object remains in the air?

### Tip:
When analyzing quadratic functions, remember that the vertex provides the maximum or minimum point, which is crucial in understanding the behavior of the function over time.

Explore how to model the motion of a dropped object using quadratic and linear functions. Learn how to calculate the height at any time using equations f(t) = –16t^2 + 250 and g(t) = 15t. Suitable for students in Grades 9-12.

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