Question

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?

Answer 1

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

Step-by-step 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.