Answer:
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<em> t</em> 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.
