SOLUTION: could you please solve the following question, a ball is thrown up into the air. its height h, in meters, after t seconds is h=-4.9t^2+38t+1.75. for what length of time is the

Question 169698: could you please solve the following question,
a ball is thrown up into the air. its height h, in meters, after t seconds is
h=-4.9t^2+38t+1.75.
for what length of time is the ball above 50 m?
and thank you,
i will be waiting for the solution,
and thank you again,

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
a ball is thrown up into the air. its height h, in meters, after t seconds is
h=-4.9t^2+38t+1.75.
for what length of time is the ball above 50 m?
----------------------------------

----------------------------------
-4.9x^2 + 38x + 1.75 = 50
-4.9x^2 + 38x - 48.25 = 0
---
x = [-38 +- sqrt(38^2 - 4*-4.9*-48.25)]/-9.8
x = 1.59973..; x = 6.1554
----------------------------
Total time at or above 50 m: 6.1554-1.59973 = 4.55564.. seconds
===============================================================
Cheers,
Stan H.

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