How Do You Calculate the Meeting Point of Two Stones Thrown from a Cliff?

[Jordan Jones]

Homework Statement

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?)

Homework Equations

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.

 

[kuruman]

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.

This is the right equation to use, but you need to write two separate equations giving the height of each stone above ground at any time t. Then you need to say that there is a specific time t_c at which the stones are at the same height. Solve for the time t_c and then use it in either equation to find the desired height.