i have a quetion for you Dr., I know this is a simple problem but my physics book is not very clear on how to figure out this answer to the question. "A stone dropped from the top of a cliff. it hits the ground below after 3.25s. how high is the Cliff?"
*to figure this out do you not use this equation? V^2 =Vo^2+ 2a(X-Xo)* a=9.8m/s, V=3.25s, initial V=0, X=?

I might point out a couple of things. First the post you have tagged is 3 years old. And it it a tutorial about kinematics in 1 dimension.

Second the problem you related suggests that the time to the bottom of the cliff is 3.25 seconds. This would not be a good thing to substitute directly as Velocity.

If you know gravity (9.8 m/s^{2}) and you know time then the correct relationship would be x = 1/2 a t^{2}