To determine how far the firework will travel before reaching the ground, we need to solve the equation \(0 = -\frac{25}{36}x^2 + \frac{50}{3}x\) (as \(16 \frac{2}{3}\) can be converted to \(\frac{50}{3}\)) for the values of \(x\) in the table.
We can set up the equation:
\[ -\frac{25}{36}x^2 + \frac{50}{3}x = 0 \]
Factoring out \(x\):
\[ x \left(-\frac{25}{36}x + \frac{50}{3}\right) = 0 \]
This gives us two possible solutions:
- \(x = 0\) (which represents when the firework is launched)
- Solving for the second factor:
\[ -\frac{25}{36}x + \frac{50}{3} = 0 \]
Multiplying through by \(-36\) to eliminate the fraction:
\[ 25x = -600 \]
Now, isolate \(x\):
\[ 25x = 600 \quad \Rightarrow \quad x = \frac{600}{25} = 24 \]
So the height of the firework reaches the ground again when \(x = 24\) feet.
Thus, the correct response from the options is:
24 feet.