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.