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 6.00m. The stones are thrown with the same speed of 9.0m/s. Find the location (above the base of the cliff) of the point where the stones cross paths.
x = 6 m
Vf = 9 m/s
a = -9.8 m/s2 (i think?)
x = (Vo)(t) + 1/2(a)(t^2) or
Vf^2 = Vo + 2(a)(x) maybe?
The Attempt at a Solution
I was thinking about using x = (Vo)(t) + 1/2(a)(t^2) but it's not working and I can't tell if i'm not using it correctly or it's just not the right equation.
6 = (0 m/s)(t) + 1/2(-9.8m/s^2)(t^2)
-1.22 = t^2
This obviously isn't going in the right direction.
Could anyone help me get started? I've understood most of the other free-fall problems in my book but the logic or strategy needed here is really confusing me.