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?
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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