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Question

A stone A is dropped from rest from height h above the ground. A second stone B is simultaneously thrown vertically up from a point on the ground with velocity v. Find v so that B meets A midway between their initial positions.

A
gh
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B
2gh
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C
3gh
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D
4gh
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Solution

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=12gt2t=hg
For stone B,h2=vt12gt2
Putting t=hg, we get
h2=vhg12g(hg)2
v=gh


Alternate Solution:
For stone B w.r.t A,
Srel=SBSA=h2(h2)=h
urel=v0=v m/s and arel=0
Using second equation of motion,
Srel=urelt+12at2
h=vt
t=hv ... (1)
For stone A:
h2=12gt2t=hg ... (2)
From (1) & (2),
hg=hvv=gh

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