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Answer to Question #145795 in Physics for Cath
Question #145795
A stone is thrown directly up at a speed of 39.2 m/s. (Show all steps) At what time does the stone reach a height of 58.8 m above where it was thrown?
Expert's answer
"h=vt-0.5gt^2\\\\58.8=39.2t-0.5(9.8)t^2\\\\t_1=2\\ s\\\\t_2=6\\ s"
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Comments
Dear Isabelle Gireada, this answer comes from the solution of a quadratic equation
could you explain how you got t1=2 s and t2=6 s?
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