Wed November 30, 2011 By: Abhinav Rastogi

A stone is dropped from a building and 2 seconds later another stone is dropped. How far apart are these two stones by the time the first one has reached at speed of 30 m/s? (Take g = 10m/s2)

Expert Reply
Wed November 30, 2011
The time taken by the first one to attain the speed 30m/s will be t =v/g = 30/10 = 3 s
So the other stone would have been in the falling only for 3-2=1 s
Now distance covered by first stone in 3 s = 45 m (s = 1/2 g t2)
The distance covered by the second one in 1 s = 5m
Hence the distance apart between the two will be 40 m.
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