Freefall Problem: Solving for the Height of a Cliff

[antimatter1422]

hi. I've been trying to do this problem for about 5 hrs...and i still can't get it! If anyone of you could please point me in the right direction, i'd really appreciate it.

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 I am 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 I am doing. several pages of random tries and still nothing... thanks in advance.

- frustrated high school student

 

[Fermat]

Using your eqn 2),
dist fallen in t secs is x
dist fallen in (t-1) secs is (2/3)x

 

[EnumaElish]

When you substitute Xo = Vo = 0 and a = g, then you have 2 Eq'ns in 2 unknowns (X and t):

X = (1/2)*a*t^2
2X/3 = (1/2)*a*(t-1)^2

 

[antimatter1422]

Relief!

thank you very much for your help Fermat and EnumaElish. I c how to work the problem now. till we meet again, bye. :rofl: