When set off, a certain firework follows the path of the quadratic function h=−2536x2+1623x

To determine how far the firework will travel before reaching the ground, we need to find the values of \( x \) for which \( h = 0 \). That means we will solve the equation:

\[ 0 = -\frac{2536}{36}x^2 + \frac{1623}{36}x \]

This is a quadratic equation in standard form \( ax^2 + bx + c = 0 \).

First, simplify the equation:

\[ 0 = -2536x^2 + 1623x \]

To solve for \( x \), we can factor out \( x \):

\[ 0 = x(-2536x + 1623) \]

Setting each factor equal to zero gives us:

1. \( x = 0 \)
2. \(-2536x + 1623 = 0\)

From the second factor, we solve for \( x \):

\[ -2536x + 1623 = 0 \ 2536x = 1623 \ x = \frac{1623}{2536} \]

Now, calculate \( \frac{1623}{2536} \):

Using a calculator or performing the division gives:

\[ x \approx 0.640 \]

Thus the firework will travel approximately \( 0.640 \) feet before hitting the ground besides the starting point at \( x = 0 \).

Conclusion: The firework travels a horizontal distance of approximately \( 0.640 \) feet before reaching the ground.