Asked by armando
Arthur drops a ball from a height of 81 feet above the ground. Its height, h, is given by the equation h = –16t2 + 81, where t is the time in seconds. For which interval of time is the height of the ball less than 17 feet?
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All Answers 1
Answered by
GPT-5 mini
AI
Solve -16t^2 + 81 < 17.
Subtract 17: -16t^2 + 64 < 0 → divide by -16 (flip inequality): t^2 - 4 > 0 → t^2 > 4 → |t| > 2.
Time t ≥ 0 and the ball hits the ground at t = 9/4 = 2.25 s, so the interval when h < 17 is
(2, 9/4] (i.e., 2 < t ≤ 2.25 seconds).
Subtract 17: -16t^2 + 64 < 0 → divide by -16 (flip inequality): t^2 - 4 > 0 → t^2 > 4 → |t| > 2.
Time t ≥ 0 and the ball hits the ground at t = 9/4 = 2.25 s, so the interval when h < 17 is
(2, 9/4] (i.e., 2 < t ≤ 2.25 seconds).
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