Help with physics!!?

Oh goody, I get to break the tie.

Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The height of the cliff is H = 6.18 m. The stones are thrown with the same speed of Uy = 8.60 m/s from the base and Vy = - 8.6 m/s from the edge.

Find the location (above the base of the cliff) of the point where the stones cross paths. That is y(U) = y(V) = ?.

y(U) = 0 + Uy T - 1/2 g T^2 = H - Vy T - 1/2 g T^2 = y(V); solve for T flight time to where they are same height y.

H - (Uy + Vy)T = 0; T = 6.18/(17.2) = ? seconds. So y(U) = y(V) = y = 8.6*6.18/(17.2) - 4.9*(6.18/(17.2))^2 = 2.46 m up from the base. ANS. Damon wins.

Let v0 = 8.6 m/s, h = height of cliff = 6.18. Let down be the negative y direction so g = -9.8 m/s^2.

Ok. The stone thrown upward has it's position as a function of time given by

y_up = 1/2 gt^2 + v0*t

while the stone thrown downward's position is

y_down = 1/2 gt^2 - v0t + h

Set these equal to each other and solve for t

y_up = y_down --> 1/2 gt^2 + v0t = 1/2 gt^2 - v0t + h

2v0t = h --> t = h/(2v0) = 6.18m/ (2*8.6m/s) = 0.359 s

Now plug t into one of the position equations to find the height above ground:

y = 1/2 gt + v0t = (1/2)(-9.8 m/s^2)(0.359s)^2 + 8.6 m/s* 0.359 s = 1.38 m