Arock dropped from a cliff covers one-third of its total distance to the ground in the last second of its fall. air resistance is negligible. how high is the cliff?

 So lets say that the total height the object falls from is expressed as: 
h = 0.5gt₁²>(1) 

Here, t₁ represents the TOTAL time the object takes to fall through h. 
Now the distance it falls through from rest until it has 1.00s to fall is: 

h - 0.33h = 0.5gt₂² 
0.667h = 0.5gt₂²>(2) 
Now that we know that it takes 1.00s to fall part of h, we can express it as: 
t₁ - t₂ = 1.00s 
t₂ = t₁ - 1.00s>(3) 

We can eliminate t₂ by subbing (1) and (3) into (2): 
0.667(0.5gt₁²) = 0.5g(t₁ - 1.00s)² 
0.3335gt₁² = 0.5gt₁² - gt₁+ 0.5g 

This is a quadratic equation, in standard form: 
0.1665gt₁² - gt₁+ 0.5g = 0 
0.1665(-9.80m/s²)t₁² - (-9.80m/s²)t₁+ 0.5(-9.80m/s²) = 0 
(-1.6317m/s²)t₁² + (9.80m/s²)t₁- 4.90m/s² = 0 
From the quadratic formula: 
t₁ = -b ± √[b² - 4ac] / 2a 
= -(9.80) ±√[(9.80)² - 4(-1.6317)(-4.90)] / 2(-1.6317) 
t₁ = 0.55s or t₁ = 5.456s 

Lets plug them both individually into (3) and if one value for t₂is negative, choose the other: 
t₂ = t₁ - 1.00s 
= 0.55s - 1.00s 
= -0.45s 

So:
t₂= 5.456s - 1.00s 
= 4.46s 

The distance fallen is easy, as you only need put the total time t₁ into (1): 

h = 0.5gt₁² 
= 0.5(9.80m/s²)(4.46s)² 
= 97.5m 

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