Motion problem? (Ball is thrown upward)

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toyotadude
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Homework Statement



Diagram of a person throwing a ball upward, with a dotted multi-exposure trajectory showing the ball rising to a maximum height h_max and returning to the release height; labels v0 and h_max are shown beside the path.


The Attempt at a Solution



I assumed that it was the same time going up as it was going down, and thus I had:

V(initial) = 0 m/s
a = -9.81 m/s^2
t = .84 seconds

And to solve for speed, I used the Distance = Speed x Time, thus Speed = Distance/Time... and used the Displacement = 1/2(V(initial) + V(final))*time to find the "distance", and then divide it by .84 seconds... but this was incorrect? ):
 
"Distance = Speed x Time" does not apply because the motion is accelerated. Only use formulas with acceleration in them. Your insight that the time to fall is equal to the time to rise may well simplify the problem - consider only the falling part.