Answer
The second rock will land 2.4s after the first rock
Explanation:
Given that
Height of the building s=50m
We assume that the first rock is acting with gravity so that a=9.81m/s
And initial velocity u=0
Applying the equation of motion
S=ut+1/2at²
50=0*t+1/2(9.81)t²
50=4.905t²
t²=50/4.905
t²=10.19
t=√10.19
t=3.19sec
For the second rock initial velocity u=8m/s and v=0 and a=9.81
Applying the equation of motion
v=u+at
0=8+9.81t
t=-8/9.81
t=0.81sec
Hence the second rock will land 2.4s after the first rock
I.e
3.19-0.81
