A toy rocket is fired from ground level. The height of the rocket with respect to time can be represented by a quadratic function. If the toy rocket reaches a maximum height of 34 feet 3 seconds after it was fired, which of the following functions could represent the height, h, of the rocket t secnds after it was fired ?

Options

A. `h (t) =-16 (t-3)^(2) + 34`
B. `h (t)=-16 (t+3) ^(2) + 34`
C. `h(t)=16 (t-3)^(2) +34`
D. `h(t) =16 (t +3) ^(2) +34`

Explanation

To solve the problem, we need to represent the height of the toy rocket as a quadratic function based on the information given. The maximum height of the rocket is 34 feet, and it occurs 3 seconds after it is fired from the ground.

### Step-by-Step Solution:

1. **Understanding the Quadratic Function**:
The height \( h(t) \) of the rocket can be modeled by a quadratic function in the standard form:
\[ h(t) = a(t - h)^2 + k \]
where \( (h, k) \) is the vertex of the parabola. Since the rocket reaches a maximum height, the parabola opens downwards, meaning \( a < 0 \).

2. **Identifying the Vertex**:
From the problem, we know that the maximum height (the vertex) is 34 feet at \( t = 3 \) seconds. Thus, we can identify:
\[ h = 3 \quad \text{and} \quad k = 34 \]

3. **Substituting into the Function**:
Plugging these values into the vertex form of the quadratic function, we have:
\[ h(t) = a(t - 3)^2 + 34 \]

4. **Determining the Value of \( a \)**:
The problem states that the rocket is fired from ground level, which means when \( t = 0 \), \( h(0) = 0 \). We can use this information to find \( a \):
\[ 0 = a(0 - 3)^2 + 34 \]
\[ 0 = 9a + 34 \]
\[ 9a = -34 \]
\[ a = -\frac{34}{9} \]

5. **Final Function**:
Substituting \( a \) back into the equation gives us:
\[ h(t) = -\frac{34}{9}(t - 3)^2 + 34 \]

6. **Choosing the Correct Option**:
We need to check which of the given options matches our derived function. The function must have a negative leading coefficient and should pass through the points we calculated.