How long after the second ball is thrown do the two balls pass each other?

[Toranc3]

Homework Statement

A juggler performs in a room whose ceiling is 3m above the level of his hands. He throws a ball upward so that it just reaches the ceiling. a) What is the initial velocity of the ball? b) What is the time required for the ball to reach the ceiling? At the instant when the first ball is at the ceiling, he throws the second ball upward with two-thirds the initial velocity of the first c) How long after the second ball is thrown did the two balls pass each other? d) What distance above the juggler's hand do they pass each other?

Homework Equations

Vy=Vo +a*t

y-yo=(voy+vy)/2 *t

y=yo+vo*t + 1/2*a*t^(2)

The Attempt at a Solution

I got A and B right.

A)
y-yo=(voy+vy)/2 *t

3m=vo/2*t

t=6m/vo

vy=vo+a*t

vo=g*t

vo=g*6m/vo

vo^(2)=g*6m

vo=7.672m/s

B) v=vo+a*t

0=7.672m/s-g*t
t=0.7820 seconds

C) This is where I am stuck at.

So it asks this: How long after the second ball is thrown do the two balls pass each other?
Is this referring to right when one is above the other or when they are equal in height? I am confused on this part. Thanks in advance for the help.

 

[mfb]

Is this referring to right when one is above the other or when they are equal in height?

Same height. In other words, they have the same value for their y-position.

 

[Toranc3]

Same height. In other words, they have the same value for their y-position.

Thanks!