Astone is dropped from the roof of a high building. a second stone is dropped 1.25 s later. how long does it take for the stones to be 25.0 meters apart?

It will take the stones 2.67s to be 25 m apart.

Explanation:

if s is the distance between the stones.

for the first stone:

s = u×t + 1/2×g×t^2

since u is the initial velocity, u = o m/s.

s1 = 1/2×g×t^2

for the second block:

s2 = 1/2×g×(t - 1.25)^2

then s1 - s2 = 25 m,

s1 - s2 = 1/2×g×t^2 - 1/2×g×(t - 1.25)^2

     25 =  1/2×g×t^2 - 1/2×g×(t - 1.25)^2

    50 =  g×t^2 - g×(t - 1.25)^2

50/g = t^2 - (t - 1.25)^2

50/g = t^2 - (t^2 - 2.5×t + 1.5625)

50/g =  2.5×t - 1.5625

       t = [50/g+1.5625]/(2.5)

          = [50/(9.8)+1.5625]/(2.5)

          = 2.67 s  

Therefore, it will take the stones 2.67s to be 25 m apart.

50 inches should be the resultant vector