A machine launches a softball from ground level. The softball reaches a maximum height of 51.84 meters above the ground...

1. TRANSLATE the problem information

• Given information:
  • Launched from ground level (\(\mathrm{h = 0}\) when \(\mathrm{t = 0}\))
  • Maximum height: 51.84 meters at \(\mathrm{t = 1.8}\) seconds
  • Hits ground: \(\mathrm{h = 0}\) when \(\mathrm{t = 3.6}\) seconds
  • Need: equation for height h at time t

2. INFER the mathematical approach

• Projectile motion follows a parabolic path (quadratic equation)
• The maximum point occurs at the vertex of the parabola
• Since we know the vertex coordinates (1.8, 51.84), look for vertex form: \(\mathrm{h = a(t - h)^2 + k}\)

3. APPLY CONSTRAINTS to test each choice

• Test each equation against all three conditions:

Choice A: \(\mathrm{h = -t^2 + 3.6}\)

• At \(\mathrm{t = 0}\): \(\mathrm{h = 3.6 \neq 0}\) ❌ (not ground level)

Choice B: \(\mathrm{h = -t^2 + 51.84}\)

• At \(\mathrm{t = 0}\): \(\mathrm{h = 51.84 \neq 0}\) ❌ (not ground level)

Choice C: \(\mathrm{h = -16t^2 - 3.6}\)

• At \(\mathrm{t = 0}\): \(\mathrm{h = -3.6 \lt 0}\) ❌ (below ground - impossible)

Choice D: \(\mathrm{h = -16(t - 1.8)^2 + 51.84}\)

• Vertex at (1.8, 51.84) ✓ (matches maximum)
• At \(\mathrm{t = 0}\): \(\mathrm{h = -16(0-1.8)^2 + 51.84}\)
  \(\mathrm{= -16(3.24) + 51.84}\)
  \(\mathrm{= 0}\) ✓
• At \(\mathrm{t = 3.6}\): \(\mathrm{h = -16(3.6-1.8)^2 + 51.84}\)
  \(\mathrm{= -16(1.8)^2 + 51.84}\)
  \(\mathrm{= 0}\) ✓

4. SIMPLIFY the verification calculations

• For Choice D at \(\mathrm{t = 0}\):
  \(\mathrm{-16(1.8)^2 + 51.84}\)
  \(\mathrm{= -16(3.24) + 51.84}\)
  \(\mathrm{= -51.84 + 51.84}\)
  \(\mathrm{= 0}\)
• For Choice D at \(\mathrm{t = 3.6}\):
  \(\mathrm{-16(1.8)^2 + 51.84}\)
  \(\mathrm{= -51.84 + 51.84}\)
  \(\mathrm{= 0}\)

Answer: D

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that the maximum height and time correspond to the vertex of a parabola in vertex form. They might focus only on testing whether equations equal zero at certain times, missing the vertex connection.

This leads them to get confused about which form to look for and may result in guessing between the remaining choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students make arithmetic errors when substituting values, particularly with the squared terms and negative coefficients in Choice D. They might incorrectly calculate \(\mathrm{-16(1.8)^2}\) or make sign errors.

This may lead them to incorrectly eliminate Choice D and select Choice A (\(\mathrm{-t^2 + 3.6}\)) or Choice B (\(\mathrm{-t^2 + 51.84}\)) thinking their calculations show D doesn't work.

The Bottom Line:

This problem requires connecting the physical meaning of "maximum height at a specific time" to the mathematical concept of a parabola's vertex, then systematically testing boundary conditions.