Help with this Physics Problem!!!! Help!!?

The general equation for height of the balls is:

h = Ho + u×t + a×t²/2

where

h = height

Ho = initial height

u = initial velocity

t = time

a = acceleration by gravity = -9.8 m/s²

The equation for displacement of the upward going ball is:

h = 0 + 25×t + (-9.8)×t²/2

h = 25t - 4.9t²

The equation for displacement of the dropped ball is:

h = 15 + 0×t + (-9.8)×t²/2

h = 15 - 4.9t²

We want the height of both balls to be the same, so

25t - 4.9t² = 15 - 4.9t²

25t = 15

t = 0.6 s < - - - - - - - - - - - - answer

Same place <--- Equation you need for each ball

At what time <--- Variable you need to solve for

Equation of motion

s = si + vi*t + 1/2*a*t^2

For this problem

h = hi + vi*t - 1/2*g*t^2

Equation for ball being dropped, vi = 0

1) h = hi - 1/2*g*t^2

h = 15 - 4.9 * t^2

Equation for the ball being thrown up... hi = 0

2) h = vi*t - 1/2*g*t^2

h = 25.0 * t - 4.9 * t^2

Since they meet at the same height, set the equations equal and solve for t

15 - 4.9 * t^2 = 25.0 * t - 4.9 * t^2

15 = 25.0 * t

t = 15 / 25 = 0.6 s