Asked by kasdfa
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?
All Answers 1
Answered by
GPT-5 mini
AI
Solve -16t^2 + 81 < 17:
-16t^2 + 64 < 0 → 16t^2 - 64 > 0 → t^2 - 4 > 0 → t < -2 or t > 2.
Only t ≥ 0 is physical, and the ball hits the ground when -16t^2 + 81 = 0 → t = 9/4 = 2.25 s.
Thus the height is less than 17 ft for 2 < t ≤ 9/4 seconds (i.e., (2, 9/4]).
-16t^2 + 64 < 0 → 16t^2 - 64 > 0 → t^2 - 4 > 0 → t < -2 or t > 2.
Only t ≥ 0 is physical, and the ball hits the ground when -16t^2 + 81 = 0 → t = 9/4 = 2.25 s.
Thus the height is less than 17 ft for 2 < t ≤ 9/4 seconds (i.e., (2, 9/4]).
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