The height of a boulder launched from a Roman catap can be described as a function of time according to the following quadratic equation: `h(t)=-16 t ^(2) +224t+240.`
How much time elapese between the moment the boulder is launched and the moment it hits the ground, assuming that the ground is at a height of 0 ?

Options

A. 7
B. 12
C. 14
D. 15

Explanation

To find the time elapsed between the moment the boulder is launched and the moment it hits the ground, we need to solve the quadratic equation given by the height function \( h(t) = -16t^2 + 224t + 240 \) when \( h(t) = 0 \).

### Step-by-Step Solution:

1. **Set the height equation to zero:**
\[ -16t^2 + 224t + 240 = 0 \]

2. **Divide the entire equation by -16 to simplify:**
\[ t^2 - 14t - 15 = 0 \]

3. **Factor the quadratic equation:**
We need two numbers that multiply to -15 and add to -14. The numbers -15 and 1 work:
\[ t^2 - 15t + t - 15 = 0 \]

4. **Group and factor by grouping:**
\[ (t^2 - 15t) + (t - 15) = 0 \]
\[ t(t - 15) + 1(t - 15) = 0 \]
\[ (t - 15)(t + 1) = 0 \]

5. **Set each factor to zero:**
\[ t - 15 = 0 \quad \text{or} \quad t + 1 = 0 \]

6. **Solve for \( t \):**
\[ t = 15 \quad \text{or} \quad t = -1 \]

7. **Interpret the results:**
Since time cannot be negative, we discard \( t = -1 \). Thus, the time when the boulder hits the ground is:
\[ t = 15 \text{ seconds} \]

### Final Answer:
The boulder hits the ground after **15 seconds**.

---