After striking the floor , a rubber ball rebounds to 4/5th of the height from which it has fallen. Find the total distance that it travels before coming to rest if it has been gently dropped from a height of 120 m.

1-540m.
2-960m.
3-1080m.
4-1020m

• {a/1-l}=={120/1-4/5+96/1-4/5}==1080
• 9 years agoHelpfull: Yes(9) No(0)
• the first drop is 120 m.After this the ball will rise by 96m and then again fall by 96 m.This process will continue in the form of infinite G.P with starting term as 120 and common ratio of 4/5 i.e. 0.8.
  So using a/(1-r)={120/(1-0.8) + 96/(1-0.8)}
  
• 9 years agoHelpfull: Yes(5) No(0)
• After the drop, it travels down for h meters and rebounds back to 4h/5 and travels down 4h/5 and rebounds back 4/5(4h/5) etc...
  
  The total distance it travels = h + 4h/5+ 4h/5+ 4/5(4h/5)+ 4/5(4h/5) + 4/5*4/5(4h/5)+...
  
  S= h + 2*4h/5 + 2(4/5)(4h/5) + ....
  
  = h + 2{4h/5 + 4/5(4h/5) + ....}
  
  = h + 2(Sum of infinite GP with initial term 4h/5 and common ratio 4/5)
  
  Formula: Sum of Infinite GP with initial term ‘a’ and common ratio ‘r’ is a/(1-r)
  
  S = h + 2{(4h/5)/(1-4/5)}
  
  = h + 8h = 9h = 9*120 = 1080
  
  Note: two things required to solve this problem are,
  
  first- deciding how much distances it travels before and after each rebound
  
  second- formula for infinite GP.
• 9 years agoHelpfull: Yes(2) No(0)
• the ans will be 1080m
• 9 years agoHelpfull: Yes(1) No(1)
• can any one explain it
• 9 years agoHelpfull: Yes(0) No(1)
• 120+ 196/(1-4/5)= 1080m
  
  Here GP formed is 196, 153.6,.....
  a=196
  r=153.6/196=4/5
  so the sum of n terms of GP is
  s=a/(1-r)
  120 is added.
• 9 years agoHelpfull: Yes(0) No(0)
• 120+2*120*(4/5(1+(4/5)+(4/5)^2+(4/5)^3+.......))
  =120+2*120*(4/5)*(1/(1-(4/5)))
  =120+2*120*4
  =1080
• 9 years agoHelpfull: Yes(0) No(0)
• can any one explain
• 9 years agoHelpfull: Yes(0) No(0)
• {120/1-4/5+90/1-4/5}=1080
• 3 years agoHelpfull: Yes(0) No(0)