Finding Initial Velocity with barely anything

[allegro1993]

Homework Statement

Someone throws a rubber ball vertically upward from the roof of a building 8.00 m in height. The ball rises, then falls. It misses the edge of the roof, and strikes the ground. If the ball is in the air for 3.00 s, what was its initial velocity? (Disregard air resistance. a = -g = -9.81m/s²)

Homework Equations

Mostly these:

\Deltay = 1/2a_y(\Deltat)²
v_{y, f} = a_y\Deltat
v_{y, f}² = 2a_y\Deltay

idk if there are equations that could be used to solve this easier.

The Attempt at a Solution

I'm completely stumped. I figured in order to find the initial velocity, I'd need to find the height of the peak. To find the height, you would need the time interval for the second half, which is after the ball reaches the peak.
All that is given time-wise is the overall time (3 seconds), and I have no idea how to find out how long it takes to reach the top of its arc.
Any help would be appreciated, I'm at a dead end.

 

[rock.freak667]

You have the equation y=y₀+ut-0.5gt²

When the ball hits the ground y=0, you have t and you want to find 'u'.

 

[allegro1993]

could you elaborate on that equation a little bit for me ?
I see what you did substituting g for a_y, that makes sense, but what does "u" stand for ?

 

[Learnphysics]

Il'l give you a start

Here it is the formula simplified

All of these values are given for the "y" axis, so don't be subbing in horizontal distance traveled or anything just vertical quantities.

displacement = Initial velocity * Time + .5 (acceleration) * Time^2



Now, the distance traveled by the ball is (8 - however high the building is.) Do we have the height of the building? No.

Can we find the height of the building?

That should give you a start on how to think of this kind of problem.

 

[allegro1993]

Il'l give you a start

Here it is the formula simplified

All of these values are given for the "y" axis, so don't be subbing in horizontal distance traveled or anything just vertical quantities.

displacement = Initial velocity * Time + .5 (acceleration) * Time^2



Now, the distance traveled by the ball is (8 - however high the building is.) Do we have the height of the building? No.

Can we find the height of the building?

That should give you a start on how to think of this kind of problem.

No, the building is 8 m high. I don't know how high the ball goes above the building :/

 

[allegro1993]

Thank you very much that helped me alot.

 

[a.ratnaparkhi]

I Know a formula. But I don't know your notations. So I'll write mine.
initial velocity= u
Height of building = h
acc. = g
time=t
-h=u*t-(1/2)gt^2
-8=3u-(1/2)9.81*9
44.14-8=3u
36.14=3u
u=12.04 m/s
is it the correct answer