The correct option is A √gh
Let time of travel of each stone =t
Distance travelled by each stone =h2
Using second equation of motion,
Taking upward as positive for both stones,
For stone A,−h2=−12gt2⇒t=√hg
For stone B,h2=vt−12gt2
Putting t=√hg, we get
h2=v√hg−12g(√hg)2
v=√gh
Alternate Solution:
For stone B w.r.t A,
Srel=SB−SA=h2−(−h2)=h
urel=v−0=v m/s and arel=0
Using second equation of motion,
Srel=urelt+12at2
⇒h=vt
⇒t=hv ... (1)
For stone A:
h2=12gt2⇒t=√hg ... (2)
From (1) & (2),
√hg=hv⇒v=√gh