- #1

Brownk

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## Homework Statement

Okay so the question is..A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 3.99999 seconds later with an initial speed of 78.3998 m/s. They hit the ground at the same time.

How long does it take the first stone to hit the ground? The acceleration of gravity is 9.8m/s^2.

and

How high is the cliff?

## Homework Equations

vf= vi + a*t

vf^2 = vi^2 +2*a*d

d = vi + (1/2)*a*(t^2)

where vf= final velocity, vi= initial velocity, a = acceleration, d = distance, t = time

## The Attempt at a Solution

Okay so this is my first time posting on this forum and forgive me if I do something wrong but here is what I tried to do.

For the first rock,

vf= ?, vi= 0m/s, t=?, a= 9.8m/s^2, and d=?

And the second,

vf=?, vi=78.3998m/s, t=?, a=9.8m/s^2, and d=?

Am I missing something or am I missing 3 variables for each of the rocks?? I don't really know where to go without another variable for one of the two...Thanks for any help