A ball is thrown upward with a height given by h(t) = -16t² + 64t + 5, where t is time in seconds. What is the maximum height?

Question

Accepted Answer

69 feet. The maximum height occurs at the vertex of the parabola. The t-coordinate of the vertex is t = -b/(2a) = -64/(2(-16)) = -64/(-32) = 2 seconds. Substituting t = 2: h(2) = -16(4) + 64(2) + 5 = -64 + 128 + 5 = 69 feet. The maximum height is 69 feet.

The value 64 is the coefficient of t (the initial velocity in feet per second) but is not the maximum height; this might result from confusing a parameter of the equation with the answer. The value 53 might come from a calculation error, perhaps incorrectly computing -16(4) as -48 or making another arithmetic mistake. The value 5 is the initial height (the constant term, representing h(0)), not the maximum height; this is where the ball starts, not where it reaches its peak.

Application problems involving quadratic functions are common on the DSST exam. Students must identify the vertex as the maximum (when a < 0) or minimum (when a > 0) point and compute it using the vertex formula. These problems appear in contexts of projectile motion, optimization, revenue, and other real-world scenarios.

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