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?
1
Expert's answer
2020-11-23T10:26:51-0500
h=vt−0.5gt258.8=39.2t−0.5(9.8)t2t1=2 st2=6 sh=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

Assignment Expert
30.11.20, 22:41

Dear Isabelle Gireada, this answer comes from the solution of a quadratic equation

Isabelle Gireada
29.11.20, 20:42

could you explain how you got t1=2 s and t2=6 s?

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