Elementary initial velocity problem

Haha guys, I have an elementary physics question I just want to make sure my understanding is correct.

Let's say I throw an object up perfectly vertical in the absence of air resistance. Time until impact back on the ground is 6 seconds. Let us also assume g = 9.80 m/s^2

Since we have no idea what displacement, Vi or Vf is, I assume the only equation I can use algebraically is

a= v/t

therefore

v = at

Substituting in my knowns I get...

v = (-9.81)(6)

However, That would return a - for my initial velocity. Almost v= at does not distinguish between vi and vf from what I can see. What am I doing wrong, or is there a resource that can guide me? Thanks guys.

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Commented:
from symmetry, Vi = -Vf
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Author Commented:
So was my calculation right minus sign?
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Commented:
As ozo said, Vi = -Vf, therefore if Vi is considered positive, your calculation yielded Vf. Does that make more sense?

Jim
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Commented:
>> So was my calculation right minus sign?

No. After 6 seconds, the object is back on the ground. That means that it took some time to reach the highest point (where its speed is 0), and some more time to reach the ground again (falling down).

So, to calculate Vi (or Vf), you know that the object decelerates from Vi to 0 in 6/2 = 3 seconds, and then accelerates from 0 to Vf also in 6/2 = 3 seconds.

Knowing that, it's easy to calculate Vi (and Vf). Just use the right values ;)
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Commented:
>> 6/2 = 3 seconds

Note that this is because of a constant acceleration.
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Author Commented:
Thanks. I kind of figured in the back of my head you divided time by 1/2, but was unsure.
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