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

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