Calculate Cliff Height: Free-Fall Problem with Rock Dropped from a Cliff

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A rock dropped from a cliff falls one-third of its total distance in the last second of its fall, prompting a calculation of the cliff's height. The problem involves using the equation d = Vot + at²/2, with initial velocity set to zero and acceleration approximated at 10/9.8 m/s². To solve, the total height is denoted as h, and time taken to fall the entire distance is represented as t + 1 seconds. Two equations are established based on the distance traveled in t seconds and t + 1 seconds, allowing for the calculation of height. The discussion emphasizes deriving these equations to find both time and height accurately.
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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?

I'm not asking for answer, I know the Physics Forum rules, but I am totally unsure where to start. I guess I have to use the equation d = Vot + at²/2 ?

Could you please give me a hand and help me figure it out, please. Hints is all I need. Thanks.
 
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lemme try

you are given initial velocity =0
acceleration is 10/9.8m/s2
let total height be h
and time taken to cover full length be t+1 sec

now find distance traveled is t seconds and then find it for t+1 sec

then u get 2 equations involving time and height

solve for answers...
 
Why t+1, though?
 
now distance traveled in last sec is known

so u can take total time as t and 1 sec before time time as t-1

or

u can take total time as t+1 and 1 sec before time as t
 
Or just solve for V_0 with d = 1/3 h and t = 1 s. Then go back and solve for the time it takes to attain that velocity in free fall.
 
h = 1/2*g*t^2 --------(1)
h - h/3 = 1/2*g*(t - 1)^2 -------(2)
Divide 1 by 2 and solve for t. Hence find h.
 
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