- #1

antimatter1422

The Problem is: A rock dropped from a cliff falls one-third of its total distance to the ground in the last second of its fall. How high is the cliff?

Answer Given = 145.7 m

I've tried drawing a diagram, substituting, but im still missing some concept! i can't get the answer given.

I know the 4 kinematic equations commonly used in class are

1. X = Xo + Vavg*t

2. X = Xo + Vo*t + (1/2)*a*t^2

3. V^2 = Vo^2 + 2*a*deltaX

4. V = Vo + a*t

i know, that since its falling off a cliff, the acceleration = g. and i have a picture but i don't know how to draw it on the computer. i've been trying to find the final velocity of the rock as it travels 2/3 down the cliff, because that final velocity would be the initial velocity of the rock as it falls the last 1/3 in a time of t= 1 second. But i can't figure out the time it takes the rock to travel the first 2/3 down the cliff. so in conclusion, i don't know what im doing. several pages of random tries and still nothing.... thx in advance.

- frustrated high school student