A ball is thrown vertically upwards to a height of 50 meter. After striking the ground it rebounds 3 by 5th of the height to which it rose first and so on for every rebound. What is the total distance traveled by the ball before it comes to rest?

• Bhaskar your answer is wrong
  first ball is thrown vertically upwards so we have add 50 to your answer since upwards 50 and while coming down 50
  so answer is 200+50=250
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• (3/5)
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• (3/5)
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• Answer is 250 -------(3/5)
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• It will first travel 50 downwards, then 3/5 of 50 = 30 upwards and the 3/5 of 50 downwards, and so on till rest.
  
  total distance can be written as,
  = 50 + (3/5)50 + (3/5)*50 + (3/5)*(3/5)*50 + (3/5)*(3/5)*50 + ......
  = 50 + 2*50 [(3/5)+ (3/5)^2 + (3/5)^3 +......]
  the terms in the bracket are in GP and it is infinite till it becomes 0
  so the sum inside the bracket can be written as
  a/(1-r) = (3/5)/(1-(3/5)) = 3/2 = 1.5 .....applying sum of infinite series.
  
  so now the total distance becomes
  = 50+2*50*1.5
  = 50+150
  = 200
  so the total distance traveled by the ball is 200 meters.
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• thanks RAM, i miss read the question.
  there is a modification in the previous answer
  i have taken that the ball dropped from that height.
  
  so we have to add 50 as, it is thrown upwards first.
  so the correct answer is 250.
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• Ans 175 50+50+30+18+10.8+6.48+3.888+2.3328+1.39968+0.839808+0.5038848.......=175 approxmatly
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