A stone is dropped from a height h. Another stone is thrown up simultaneously from the ground, which can reach to maximum height of 4h. The two stones will cross each other after time
A
√h8g
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B
√8gh
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C
√2gh
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D
√h2g
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Solution
The correct option is A√h8g Initial velocity of 2nd stone.
Using v2=u2+2as 0=u2−2g(4h) ⇒u2=√8gh
Initial velocity of first stone is u1=0
Relative velocity u21=u2−u1=√8gh
Initial relative distance between stones is drel=h
Relative acceleration between stones is arel=−g−(−g)=0
Using t=drelurel t=h√8gh=√h8g