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A body is dropped from certain height and another body is dropped from same height after 2 secs. What will be the separation between two bodies 4 second after the dropping of second body.
You can use Newton's second equation of motion here.The first ball has traveled for 6 seconds under gravity with acceleration g and initial velocity=0.The second ball has traveled for 4 seconds under gravity with acceleration g and initial velocity=0.Distance between the 2 balls at the end of 6 seconds after the first ball is dropped = 1/2*9.8(36-16)= 98m
98 meters is right ansH1=1/2×9.8×6×6H1=176.4Now,H2=1/2×9.8×4×4 H2=78.4Thus,the seperation between two bodies are H=H1-H2H=178.4-78.4H=98 meters
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