A stone 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.