Free falling object with gravity

I was trying to do this problem, but I got t negative, so I went back from the beginning and looked for my errors, but I couldn't find one. Can anyone tell me what I did wrong ? Thanks

You drop a ball from a window on an upper floor of a building and it is caught by a friend on the ground when the ball is moving with speed Vf. You now repeat the drop, but you have friend on the street below throw another ball upward at speed Vf exactly at the same time that you drop your ball from the window. The two balls are initially separated by 28.7m. At what time do they pass each other ?

Take upwards as positive, yo = 0 will be ground, H will be the distance when two balls pass each other.

For the ball going downward:
Vf = Vo - gt
Vf = - gt

-(28.7 - H) = -1/2 gt^2
H = - 1/2gt^2 + 28.7 (1)

For the ball going upwards

H = Vf - 1/2gt^2
H = - gt - 1/2 gt^2 (2)

From (1) and (2)

-1/2gt^2 + 28.7 = - gt - 1/2 gt^2

28.7 = -gt

t = - 28.7/g --------> negative ?????

Your initial steps as under are incorrect.
For the ball going downward:
Vf = Vo - gt
Vf = - gt
For the ball going downwards, Vf is in the downward direction which you have considered +ve. Hence "g" in the downward direction must be taken positive.
In all such problems, first decide which direction is to be considered +ve and which -ve and use that convention allthroughout.
In the step,
H = Vf - (1/2)gt^2, if t is the total time of travel, then error is
H = Vf * t - (1/2)gt^2.

I would prefer to solve the problem rather than checking your solution as in your solution you are confused about "t". We need to find the time when both will meet and not the time for any to cover the distance 28.7 m.

SOLUTION:

Let them meet after time "t" from the start.

First we find the value of Vf
Vf^2 - 0 = 2g*28.7
=> Vf = √[2 * (9.81) * (28.7)] = 23.7 m/s