Question:The equation \(\mathrm{h(t) = -8t^2 + 36t + 5}\) can be used to model the height, h, in meters, of...
GMAT Advanced Math : (Adv_Math) Questions
The equation \(\mathrm{h(t) = -8t^2 + 36t + 5}\) can be used to model the height, \(\mathrm{h}\), in meters, of a ball \(\mathrm{t}\) seconds after it is thrown upwards from a platform, where \(\mathrm{t \geq 0}\). The function models the ball's height from when it is thrown until it hits the ground. For what value of \(\mathrm{t}\) does the ball reach its maximum height?
1. TRANSLATE the problem information
- Given information:
- Height function: \(\mathrm{h(t) = -8t^2 + 36t + 5}\)
- This models a ball's height t seconds after being thrown
- What this tells us: We have a quadratic function in standard form where \(\mathrm{a = -8, b = 36, c = 5}\)
2. INFER the approach
- Since the coefficient of t² is negative (\(\mathrm{a = -8 \lt 0}\)), this parabola opens downward
- A downward-opening parabola has a maximum point at its vertex
- We need to find the t-coordinate of the vertex using the vertex formula
3. SIMPLIFY using the vertex formula
- For any quadratic \(\mathrm{y = at^2 + bt + c}\), the x-coordinate of the vertex is: \(\mathrm{x = -b/(2a)}\)
- Substitute our values: \(\mathrm{t = -36/(2(-8))}\)
- Calculate: \(\mathrm{t = -36/(-16) = 36/16}\)
- Reduce the fraction: \(\mathrm{36/16 = 9/4}\)
- Convert to decimal: \(\mathrm{9/4 = 2.25}\)
Answer: \(\mathrm{t = 9/4}\) seconds or \(\mathrm{t = 2.25}\) seconds
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students don't recognize this as a vertex problem. They might try to find when \(\mathrm{h(t) = 0}\) (when ball hits ground) instead of finding the maximum height. They set up \(\mathrm{-8t^2 + 36t + 5 = 0}\) and solve for t, getting completely different values that represent when the ball is at ground level, not maximum height.
This leads to confusion about what the problem is actually asking for.
Second Most Common Error:
Inadequate SIMPLIFY execution: Students correctly identify they need the vertex formula but make arithmetic errors. They might calculate \(\mathrm{-36/(2(-8))}\) incorrectly, getting \(\mathrm{-36/16}\) instead of \(\mathrm{36/16}\), leading to a negative time value of \(\mathrm{-9/4}\). Since time can't be negative in this context, this causes them to doubt their approach and guess.
The Bottom Line:
The key challenge is recognizing that "maximum height" means finding the vertex of a parabola, not solving when the height equals zero. Once students make this connection, the algebra is straightforward.