Question

a rock 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?

Answer 1

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