Stone is dropped from the top of a cliff

[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.

 

[quantumdude]

Originally posted by Alice
A stone is dropped from the top of a cliff,

That means that v_i=0, and the y_i the height of the cliff, which is unknown. Implicit in this is that the acceleration of the stone is that of gravity.

it is seen to hit the ground below after 3.66 s,

That means that Δt=3.66s. Implicit in that is that y_f=0, which is ground level.

how high is the cliff? Anyone know?

Can you find an equation that relates those quantities?

 

[svtec]

Mentor Edit: Please don't post complete solutions. Thank you.[/color]

 

[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

 

[turin]

Originally posted by Alice
/\X=X-Xo=VoxT+1/2AxT^2

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

 

[quantumdude]

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

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.