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Question: A fireworks shell is launched straight upward from a 64-foot platform with an initial vertical velocity of 160 feet...

GMAT Advanced Math : (Adv_Math) Questions

Source: Prism
Advanced Math
Nonlinear functions
HARD
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Notes
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Question:

A fireworks shell is launched straight upward from a 64-foot platform with an initial vertical velocity of 160 feet per second. The height of the shell above ground \(\mathrm{t}\) seconds after launch is modeled by \(\mathrm{h(t) = -16t^2 + 160t + 64}\), where \(\mathrm{h(t)}\) is measured in feet. What is the maximum height, in feet, that the shell reaches above ground?

Answer Format Instructions: Enter your answer as an integer.

Answer Type: Grid-in

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Solution

1. TRANSLATE the problem information

  • Given information:
    • Height function: \(\mathrm{h(t) = -16t^2 + 160t + 64}\)
    • Need to find: maximum height above ground

2. INFER the mathematical approach

  • Since \(\mathrm{h(t)}\) is a quadratic with \(\mathrm{a = -16 \lt 0}\), the parabola opens downward
  • The maximum value occurs at the vertex of the parabola
  • We need the vertex formula to find when the maximum occurs, then substitute to find the height

3. SIMPLIFY using the vertex formula

  • For quadratic \(\mathrm{at^2 + bt + c}\), the vertex occurs at \(\mathrm{t = -b/(2a)}\)
  • Here: \(\mathrm{t = -160/(2 \times (-16)) = -160/(-32) = 5}\) seconds

4. SIMPLIFY by substituting to find maximum height

  • \(\mathrm{h(5) = -16(5)^2 + 160(5) + 64}\)
  • \(\mathrm{h(5) = -16(25) + 800 + 64}\) (use calculator)
  • \(\mathrm{h(5) = -400 + 800 + 64 = 464}\) feet

Answer: 464




Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize they need to find the vertex of a parabola. Instead, they try to solve \(\mathrm{h(t) = 0}\) to find when the shell hits the ground, completely missing that the question asks for maximum height. This leads to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly identify they need the vertex but make arithmetic errors, particularly when calculating \(\mathrm{t = -b/(2a) = -160/(-32)}\), getting the sign wrong or miscalculating to get \(\mathrm{t = -5}\) or \(\mathrm{t = 10}\). This leads to substituting the wrong t-value and getting an incorrect maximum height.

The Bottom Line:

This problem tests whether students can distinguish between "when does something happen" (vertex t-coordinate) versus "what is the value when it happens" (vertex height). The key insight is recognizing that maximum height problems for quadratic motion always involve finding the vertex of a downward-opening parabola.

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