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Q. A STONE IS THROWN UPWARDS WITH AN INITIAL VELOCITY v0. THE DISTANCE TRAVELLED IN TIME 4v0/3g IS-
  1. 2v02/g
  2. v02/2g
  3. 4v02/3g
  4. 5vo2/9g

Bharat Makkar , 11 Years ago
Grade Select Grade
anser 1 Answers
shashi K Sharma
Dear Bharat
The question asks of distance and you must be careful about it as the body changes its direction after maximum height(time=v/g)
Thus the body moves up till time = u/g & then down for rest time (4v/3g – v/g = v/3g =t’ say)
Thus the distance is maximum height plus the distance moved down.
the maximum height being (v2)/2g &  the distance fallen in the aforesaid time t’ is ½gt’2
which comes as   v2/18g. Thus the sum total of the two is 5v2/9g
Last Activity: 11 Years ago
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