Question

When set off, a certain firework follows the path of the quadratic function h= -25/36x^2 + 16(2/3x)

X= the horizontal distance it travels
H= the height of the firework in feet

Determine how far the firework will travel before reaching the ground, determine the value that is the solution to the equation 0= -25/36x^2 + 16(2/3x)

Answer 1

To determine how far the firework travels before reaching the ground, we need to find the value of \( x \) when the height \( h \) is equal to zero. We can set the quadratic function equal to zero:

\[
0 = -\frac{25}{36}x^2 + 16\left(\frac{2}{3}x\right)
\]

First, simplify the second term:

\[
16\left(\frac{2}{3}x\right) = \frac{32}{3}x
\]

Now the equation becomes:

\[
0 = -\frac{25}{36}x^2 + \frac{32}{3}x
\]

Next, multiply the entire equation by 36 to eliminate the fraction:

\[
0 = -25x^2 + 12 \cdot 32x
\]

Now calculate \( 12 \cdot 32 \):

\[
0 = -25x^2 + 384x
\]

Rearranging gives us a standard quadratic form:

\[
25x^2 - 384x = 0
\]

Next, factor out \( x \):

\[
x(25x - 384) = 0
\]

Setting each factor equal to zero gives us the solutions:

1. \( x = 0 \)
2. \( 25x - 384 = 0 \)

Solving the second equation for \( x \):

\[
25x = 384
\]
\[
x = \frac{384}{25}
\]
\[
x = 15.36
\]

Thus, the firework will travel **15.36 feet** before reaching the ground. The solutions to the equation \( 0 = -\frac{25}{36}x^2 + 16\left(\frac{2}{3}x\right) \) are \( x = 0 \) and \( x = 15.36 \).