# Finding time for a particular position of vertically thrown baseball

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1. Jul 12, 2016

### Zack K

1. The problem statement, all variables and given/known data
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

2. Relevant equations
Vf=Vi+at
d=Vit+½at2

3. 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.

2. Jul 12, 2016

### 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.

3. Jul 12, 2016

### 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.

4. Jul 12, 2016

### Zack K

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