Two stones are dropped from the edge of a 60m cliff , the second stone 1.6secon after the first . How far below the top of the cliff is the second stone when the separation between the two stone is 36m?

The separation between the two stones is 36 m, when the second stone is approximately 10.9 m below the top of the cliff

Explanation:

The given parameters are;

The height of the cliff from which the stones are dropped, h = 60 m

The time at which the second stone is dropped = 1.6 seconds after the first

The distance below the top of the cliff when the distance between the two stones is 36 m = Required

We have;

The kinematic equation of motion that can be used is s = u·t - (1/2)·g·t²

For the first stone, we have, s₁ = u·t₁ - (1/2)·g·t₁²

For the second stone, we get; s₂ = u·t₂ - (1/2)·g·t₂²

t₁ = t₂ + 1.6

g = The acceleration due to gravity ≈ 9.81 m/s²

s = The distance below the cliff top

The initial velocity of the stones, u = 0

Let t represent the time from which the second stone is dropped at which the distance between the two stones is 36 m, we have;

s₁ = u·(t + 1.6) + (1/2)·g·(t + 1.6)²

s₂ = u·t + (1/2)·g·t²

u = 0

∴ s₁ - s₂ = 36 =  (1/2)·g·(t + 1.6)² - (1/2)·g·t²

2 × 36/(g) = (t + 1.6)² - t²  = t² + 3.2·t + 2.56 - t² = 3.2·t + 2.56

2 × 36/(9.81) = 3.2·t + 2.56

t = (2 × 36/(9.81) - 2.56)/3.2 =  ≈ 1.49 s

t ≈ 1.49 s

s₂ = (1/2)·g·t²

∴ s₂ = (1/2) × 9.81 × 1.49² ≈ 10.9

The distance below the top of the cliff of the second stone when the the separation between the two stones is 36 m, s₂ ≈ 10.9 m.

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there are no contact forces mentioned on the list of choices that you've provided.