SOLUTION: When a ball is thrown upward, its height, h meters, is given by h=1.5+19.6t-4.9t2, where t seconds is the time after it is thrown. For what length of time is the ball higher than 1

Question 1199441: When a ball is thrown upward, its height, h meters, is given by h=1.5+19.6t-4.9t2, where t seconds is the time after it is thrown. For what length of time is the ball higher than 16.2m?
Found 2 solutions by ikleyn, Alan3354:
Answer by ikleyn(52778) (Show Source): You can put this solution on YOUR website!
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When a ball is thrown upward, its height, h meters, is given by h=1.5+19.6t-4.9t2,
where t seconds is the time after it is thrown.
For what length of time is the ball higher than 16.2m?
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To answer the question, you need to solve this inequality 1.5 + 19.6t - 4.9t^2 > 16.2. Transform it to an equivalent inequality 4.9t^2 - 19.6t + 14.7 < 0 Divide both sides by 4.9 t^2 - 4t + 3 < 0. Factor left side (t-3)*(t-1) < 0 The solution set is 1 < t < 3. The ball will be above 16.2 m for 3 - 1 = 2 seconds. ANSWER

Solved.

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
When a ball is thrown upward, its height, h meters, is given by h=1.5+19.6t-4.9t2, where t seconds is the time after it is thrown. For what length of time is the ball higher than 16.2m?
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The 2 seconds answer from Ikleyn is the time it's at or above 16.2 meters, not above.
So it's 2 - dt seconds.
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dt = Planck time seconds
Maybe 2 Planck times.
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