A ball is thrown upward, and its height above the ground is modeled by a quadratic function \(\mathrm{h(t)}\), where t...

1. TRANSLATE the problem information

• Given information:
  • Ball's height follows quadratic function \(\mathrm{h(t)}\)
  • Maximum height 20 feet at \(\mathrm{t = \frac{5}{2}}\) seconds (vertex)
  • Ball hits ground (\(\mathrm{h = 0}\)) at \(\mathrm{t = \frac{9}{2}}\) seconds (one root)
  • Need to find the other ground contact time (other root)

2. INFER the key relationship

• Since this is a quadratic function, the graph is a parabola
• The vertex at \(\mathrm{t = \frac{5}{2}}\) represents the axis of symmetry
• Due to parabolic symmetry, the two roots (where ball hits ground) must be equidistant from the vertex

3. SIMPLIFY using symmetry properties

• Distance from vertex to known root: \(\mathrm{\frac{9}{2} - \frac{5}{2} = \frac{4}{2} = 2}\) seconds
• The other root must be the same distance on the opposite side
• Other root = \(\mathrm{\frac{5}{2} - 2 = \frac{5}{2} - \frac{4}{2} = \frac{1}{2}}\) second

Answer: B (\(\mathrm{t = \frac{1}{2}}\) second)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize the symmetry property of parabolas. They might try to create the full quadratic equation using all three points, leading to unnecessary complex calculations. This causes them to get stuck and abandon systematic solution, often guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students understand the symmetry concept but make arithmetic errors with fractions. For example, calculating \(\mathrm{\frac{5}{2} - 2}\) incorrectly as \(\mathrm{\frac{3}{2}}\) instead of \(\mathrm{\frac{1}{2}}\), leading them to select Choice C (\(\mathrm{t = \frac{3}{2}}\) seconds).

The Bottom Line:

This problem tests whether students can recognize and apply the fundamental symmetry property of parabolas rather than getting bogged down in complex algebraic manipulations. The key insight is that quadratic functions create symmetric graphs, making this a pattern recognition problem disguised as a complex calculation.