Question 21 • Mathematics III
The object is in the air from launch at t = 0 until its height returns to 0. Solving −4.87t² + 18.75t = 0 gives t = 0 or t = 18.75 ÷ 4.87 ≈ 3.85, so the domain is [0, 3.85]. The parabola opens downward, and its maximum height occurs at t = −18.75 ÷ (2 × −4.87) ≈ 1.92 seconds. Substituting this time gives a maximum height of about 18.05 meters, so the range is [0, 18.05], matching choice A.
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