To determine how far the firework will travel before reaching the ground, we need to find the value of \( x \) that makes the function \( h = -\frac{25}{36}x^2 + \frac{1623}{100}x = 0 \).
This corresponds to finding the roots of the quadratic equation:
\[ -\frac{25}{36}x^2 + \frac{1623}{100}x = 0 \]
We can factor out \( x \):
\[ x\left(-\frac{25}{36}x + \frac{1623}{100}\right) = 0 \]
One solution is \( x = 0 \) (when the firework is launched). To find the other solution, we need to solve:
\[ -\frac{25}{36}x + \frac{1623}{100} = 0 \]
Rearranging gives us:
\[ -\frac{25}{36}x = -\frac{1623}{100} \]
Multiply through by -1:
\[ \frac{25}{36}x = \frac{1623}{100} \]
Now, we can solve for \( x \):
\[ x = \frac{1623}{100} \cdot \frac{36}{25} \]
Calculating this step-by-step:
- Compute \( \frac{1623 \cdot 36}{100 \cdot 25} = \frac{1623 \cdot 36}{2500} \).
Now, let's perform the multiplication:
\[ 1623 \times 36 = 58428 \]
and now divide:
\[ x = \frac{58428}{2500} \approx 23.3712 \]
Since this value doesn't exactly match the numbers in our table, let's evaluate the values provided:
- For \( x = 12 \):
\[ h = -\frac{25}{36}(12^2) + \frac{1623}{100}(12) = -\frac{25}{36}(144) + \frac{1623}{100}(12) \]
- For \( x = 24 \):
\[ h = -\frac{25}{36}(24^2) + \frac{1623}{100}(24) \]
- For \( x = 36 \):
\[ h = -\frac{25}{36}(36^2) + \frac{1623}{100}(36) \]
- For \( x = 48 \):
\[ h = -\frac{25}{36}(48^2) + \frac{1623}{100}(48) \]
We can find which values produce 0. However, as expected from the derivations:
The approximate value we calculated for \( x \) is indeed closest to 24 feet, as it is near to 23.3712.
Thus, the answer to how far the firework will travel before reaching the ground is:
24 feet.