# Falling Object

1. Feb 12, 2004

### Alice

Hey

This seems to be way easy, but i'm just not gettng where to start it. A stone is dropped from the top of a cliff, it is seen to hit the ground below after 3.66 s, how high is the cliff? Anyone know? Thanks.

2. Feb 12, 2004

### Tom Mattson

Staff Emeritus
That means that vi=0, and the yi the height of the cliff, which is unknown. Implicit in this is that the acceleration of the stone is that of gravity.

That means that &Delta;t=3.66s. Implicit in that is that yf=0, which is ground level.

Can you find an equation that relates those quantities?

3. Feb 12, 2004

### svtec

Mentor Edit: Please don't post complete solutions. Thank you.

Last edited by a moderator: Feb 12, 2004
4. Feb 12, 2004

### Alice

Tom-

I pretty much got that far on my own. I just don't know which formula ot use. The one that looks the closes would be

/\=Delta

/\X=X-Xo=VoxT+1/2AxT^2

and

Vx^2-Vox^2=2Ax/\x

but i'm not sure. I'm having a really hard time figuring out which formula goes where. Thanks.

-Alice

5. Feb 12, 2004

### turin

This one always works for a constant acceleration. And, of course, that "X" doesn't have to be horizontal.

6. Feb 13, 2004

### Tom Mattson

Staff Emeritus
Right, the equation also holds for "y".

Alice, look at the equation that turin referred to, rewrite it for y (just replace x with y), and verify that you do in fact have all the information to solve the problem. Then, plug in the numbers.