When set off, a certain firework follows the path of the quadratic function h=-25/36x62 + 16 2/3x, where :h = the height of the firework in feet and x=the horizontal distance in travels in feet

To determine which value of \( x \) corresponds to the firework reaching the ground, we need to solve the equation:

\[ 0 = -\frac{25}{36}x^2 + \frac{50}{3}x \]

This simplifies to solving:

\[ \frac{25}{36}x^2 - \frac{50}{3}x = 0 \]

Factoring out \( x \):

\[ x \left(-\frac{25}{36}x + \frac{50}{3}\right) = 0 \]

Setting each factor to zero, we have:

1. \( x = 0 \)
2. \(-\frac{25}{36}x + \frac{50}{3} = 0\)

Let's solve the second equation:

\[ -\frac{25}{36}x + \frac{50}{3} = 0 \]

Rearranging gives:

\[ \frac{25}{36}x = \frac{50}{3} \]

Multiplying both sides by \( \frac{36}{25} \):

\[ x = \frac{50}{3} \cdot \frac{36}{25} \]

Calculating the right-hand side:

\[ x = \frac{50 \times 36}{3 \times 25} = \frac{1800}{75} = 24 \]

Thus, \( x = 24 \) is the value where the firework will reach the ground.

Checking the table’s values (12, 24, 36, 48), the solution to the equation is:

\[ \boxed{24} \]