To determine how far the balloon will travel before it reaches the ground, we need to set the height \( h(x) \) to zero and solve for \( x \).
The path of the balloon is given by the equation:
\[ h(x) = x^2 + 12x \]
Setting \( h(x) \) equal to zero, we have:
\[ 0 = x^2 + 12x \]
This quadratic equation can be factored as follows:
\[ 0 = x(x + 12) \]
Setting each factor to zero gives us two possible solutions:
- \( x = 0 \)
- \( x + 12 = 0 \) → \( x = -12 \)
Since we are looking for the distance the balloon travels before it reaches the ground, we discard the negative solution.
Thus, the distance \( x = 0 \) represents the point where the balloon is fired, and as the balloon falls back to the ground after reaching a height, it will not travel further than where it was launched. However, since the opening of the given problem mentions "the balloon travels before it reaches the ground," we should check for any additional interpretations or possible distance metrics.
To find the point at which the balloon reaches the ground, we can also consider if we were to find \( x \) at the vertex of the parabola defined by \( h(x) \), but here it primarily traverses directly back to the ground as denoted.
Thus, the balloon will travel 0 feet horizontally before returning to ground level assuming we are interpreting ground-level hits since it is launched strictly downward.
If considering that it will drop directly back down to the place of launch, thus effectively, the distance it travels horizontally at the height would effectively be 0 as it returns essentially backtracking.
Based on the content interpretation: The balloon will travel 0 feet before it reaches the ground.
Please confirm if looking for an evaluation or vertex calculation scenario extrapolated or further depth required besides grounding evaluation.
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