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please answer this

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Asked by Prashant DIGHE 29th January 2020, 10:20 PM
Answered by Expert
Answer:
Let t seconds be the time of collision after the particles started moving from top and bottom.
 
Vertical distance h1 travelled by particle which is dropped from the height h is given by,
 
h1 = (1/2)gt2 .....................(1)
 
Vertical distance h2 travelled by the partcle which is projected from ground with speed begin mathsize 14px style square root of 2 space g space h end root end style is given by,
h2 = (begin mathsize 14px style square root of 2 space g space h end root end style) t - [ (1/2) g t2 ]  ..................(2)
from eqn.(1) and (2),  h1 + h2 = h = (begin mathsize 14px style square root of 2 space g space h end root end style) t  .....................(3)
hence time of collision from starting time  is given by,  begin mathsize 14px style t space equals space square root of fraction numerator h over denominator 2 g end fraction end root end style  ...................(4)
hence from eqn.(1) and eqn.(4), we get  h1 = (1/2)g (h/2g) = (1/4)h
 
Hence collision has taken place at a height (3/4)h above ground level
 
Since collision is completely inelastic, all the kinetic energy is lost and the particles stick together.
 
Hence after collision, the combined partcles start from rest to reach ground .
 
If time taken is τ for the combined particle to reach ground , then we have , (3/4) h = (1/2) g τ2
 
hence time begin mathsize 14px style tau space equals space square root of 3 over 2 end root space open parentheses square root of h over g end root close parentheses end style   ;  hence time τ in units of begin mathsize 14px style square root of h over g end root end style is begin mathsize 14px style square root of 3 over 2 end root space end style
Answered by Expert 30th January 2020, 8:56 PM
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