The correct option is A √gh
Let time of travel of each stone =t
Distance travelled by each stone =h2
Taking upward direction as positive
Using second equation of motion,
For stone A, −h2=−12gt2
⇒t=√hg
For stone B, h2=vt−12gt2
Putting t=√hg, we get
v=√gh
Alternate sol:
For stone B w.r.t A,
Srel=h, urel=v−0=v m/s, arel=0
Using second equation of motion,
Srel=urelt+12at2 ⇒h=vt ⇒t=hv……(1)
For the body dropped from height h,
h2=12gt2⇒t=√hg……(2)
From (1) and (2) √hg=hv⇒v=√hg