# Finding velocity (having trouble interpreting the question

1. Jan 22, 2015

### zx636r

1. The problem statement, all variables and given/known data
For an initial velocity of V0, what is the velocity of the ball as it passes the point A on its way up, where OA is 1/3 the maximum height.

V0 = 10m/s
t = 1s (2s total for the ball to come back down)
maximum height = 5m
initial displacement x0 = 0m

2. Relevant equations
Vf^2 = V0^2 + 2a(x - x0)

3. The attempt at a solution

I'm assuming point O is the ground. If point OA is 1/3 of the maximum height then, it is 5/3m. My confusion is what point A should be. Is it some arbitrary point that is just deemed "A"? The answers are in terms of vA (velocity as the ball passes through point A) = initial velocity times a constant. The other thing I can think of is that I am wrong for setting point O at 0. Am I supposed to go another "O" distance from OA to get to point A?

I know this answer is yahoo answers or something, but I refrained from just looking at the answer because I really want to understand what is going on.

2. Jan 22, 2015

### Bystander

You're shown a parabolic path/trajectory. "O" is at ground level, or y = 0 describes a line "O" which you could regard as the "x-axis."
Think "y = A."
Call it an "arbitrary" altitude --- "A" for "arbitrary" or for "altitude" or for "ambiguous" statement.

3. Jan 22, 2015

### Staff: Mentor

Correct.
No. According to the information you gave it is at x = 5/3 m.

Is this supposed to be solved algebraically, or is the answer supposed to be for the specific problem of v0 = 10m/s, max ht = 5 m?

No. These are both correct interpretations.

If you are trying to do this algebraically, then you use your equation to determine the maximum height x in terms of v0. To do that, you need to select the gravitational value of the acceleration a.

Chet