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a body released from a great height falls freely towards earth.another body is released from the same height exactly one second later.then the seperation between two bodies after release of two second body is
Displacement of the first body after 3 seconds= S = ut + ½ g t^2 = 44.1m/sDisplacement of the first body after 2 seconds= S = ut + ut + ½ g t^2 = 19.6m/sThe separation between them=44.5 - 19.6 = 24.5m/s
Displacement of the first body after 3 seconds= S = ut + ½ g t^2 = 44.1m/s
Displacement of the first body after 2 seconds= S = ut + ut + ½ g t^2 = 19.6m/s
The separation between them=44.5 - 19.6 = 24.5m/s
Dear Student,Please find below the solution to your problem.Displacement of the first body after 3 seconds =S = ut + ½ g t^2= 44.1m/sDisplacement of the first body after 2 seconds=S = ut + ½ g t^2= 19.6m/sThe separation between them= 44.5 - 19.6= 24.5m/sThanks and Regards
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