After striking the floor, a ball rebounces 45th of its height from which it has fallen. If it is dropped from a height of 120m, the total distance it travels before coming to rest is

• There's a theorem that states:
  The infinite sum = a/(1-r)
  a = the starting number, in this case 120.
  r = the change ration, again, in this case 1/45.
  But, this only accounts for half of the travel, we need to take in account the bounce back. So, we multiply it by 2.
  So, we have:
  2a/(1-r) = 2(120)/(1-1/45) = 245.45
  But, since the intial drop (120) isn't met twice, we need to subtract this from our answer above.
  So,
  245.45m - 120 = 124.95m
  
  
• 12 years agoHelpfull: Yes(0) No(0)
• The infinite sum in GP = a/(1-r)
  a = the starting number
  r = the common ratio in this case 1/45
  
  IT FALLSS FROM 120M.
  AFTER THAT 2(120/45+120/ 45^2+120^45^3+.........)
  =120*2(1/45+1/45^2+........)
  =240(1/45/(1-1/45))(sun of Infinite terms)
  =5.455
  
  TOTALLY IT TRAVELS 120+5.455=125.455M
  
• 12 years agoHelpfull: Yes(0) No(0)
• sorry in the first solution posted by me there is sligtly mistake in subtraction. this is correct solution after modification of that.
  
  
  
  There's a theorem that states:
  The infinite sum = a/(1-r)
  a = the starting number, in this case 120.
  r = the change ration, again, in this case 1/45.
  But, this only accounts for half of the travel, we need to take in account the bounce back. So, we multiply it by 2.
  So, we have:
  2a/(1-r) = 2(120)/(1-1/45) = 245.45
  But, since the intial drop (120) isn't met twice, we need to subtract this from our answer above.
  So,
  245.45m - 120 = 125.45m
  
• 12 years agoHelpfull: Yes(0) No(0)