2 balls are droped from same point after an interval of 1 second its acceleration due to gravity is 10 m/s2 what will be the sepration 3 second after the release first ball
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4
You can use Newton's second equation of motion here.
The first ball has traveled for 3 seconds under gravity with acceleration g and initial velocity=0.
The second ball has traveled for 2 seconds under gravity with acceleration g and initial velocity=0.
Distance between the 2 balls at the end of 3 seconds after the first ball is dropped = 1/2*9.8(9-4)
= 24.5m
The first ball has traveled for 3 seconds under gravity with acceleration g and initial velocity=0.
The second ball has traveled for 2 seconds under gravity with acceleration g and initial velocity=0.
Distance between the 2 balls at the end of 3 seconds after the first ball is dropped = 1/2*9.8(9-4)
= 24.5m
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The following image will provide the solution:
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