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Question

A stone A is dropped from rest from height h above ground. A second stone B is simultaneously thrown vertically up with velocity v. Find v so that B meets midway between their initial positions.
  1. gh
  2. 2gh
  3. 3gh
  4. 4gh


Solution

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=vt12gt2
Putting t=hg, we get
v=gh

Alternate sol:
For stone B w.r.t A,
Srel=h, urel=v0=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=12gt2t=hg(2)
From (1) and (2) hg=hvv=hg

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