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## 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.