Finding time for a particular position of vertically thrown baseball

[Zack K]

Homework Statement

A baseball pitcher makes a big mistake and throws the ball straight up. If it reaches a maximum height of 15 m. Find the times when the ball was 8.0 m above the ground.
I already solved the variables needed
Vi= 17 m/s
Time the ball spent in the air= 3.5 s
average velocity= 8.6 m/s
balls velocity when its displacement equals to 8.0= ±12 m/s
a= -9.8 m/s²

Homework Equations

Vf²=vi²+2ad
Vf=Vi+at
d=Vit+½at²

The Attempt at a Solution

I managed to find the first time by finding the balls velocity when its first 8.0 m from the ground by using the equation Vf²=vi²+2ad then plugging that into Vf=Vi+at to get a time of 0.56 s. But I just can't seem to be able to find the time for when the ball comes back down and crosses the 8.0 m mark again.

 

[TSny]

When solving Vf²=vi²+2ad for Vf, you take a square root. But this gives two possible answers.
For example, the two solutions of x² = 4 are x = 2 and x = -2.

 

[Zack K]

Well the answer is a positive because I was solving it for when the ball was hitting the 8.0 m mark while going up. But it's the coming back down part I have trouble with.

 

[Zack K]

When solving Vf²=vi²+2ad for Vf, you take a square root. But this gives two possible answers.
For example, the two solutions of x² = 4 are x = 2 and x = -2.

I managed to get the answer. Thanks for helping :)