Physics question of 1-d motion of two objects with constant acceleration.?

Question

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

Whome · Accepted Answer

using the position equation
 y = y0 + v0t + ½gt²

as initial position and velocity of each stone could be zero with a properly selected origin.
 
y = ½gt²
 
so the first stone equation is 
y1 = ½gt²
 
and the second stone equation is
 y2 = ½g(t - 1.6)²
 
so the difference is 
Δy = y1 - y2
 Δy = ½gt² - ½g(t - 1.6)² 
36 = ½g(t² - (t - 1.6)²)
 36(2)/g = (t² - (t - 1.6)²)
 72/g = t² - (t² - 3.2t + 2.56)
 72/g = 3.2t - 2.56 
72/g + 2.56 = 3.2t
 72/9.81 + 2.56 / 3.2 = t
 t = 3.09 s 

at 3.09 seconds after the first stone, the second stone has traveled
 
d = ½g(t)²
 d = ½g(3.09 - 1.6)²
 d = ½9.81(1.49)²
 d = 10.9 m

Physics question of 1-d motion of two objects with constant acceleration.?

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