math
posted by Fidelia on .
5. A rubber ball dropped on a hard surface takes a sequence of bounces, each one 3/5 as high as the preceding one. If this ball is dropped from a height of 15 feet, what is the total distance it has traveled after it hits the surface the fifth time?
The answer is 54 and 21/125 ft, but why?
A rubber ball dropped on a hard surface takes a sequence of bounces, each one 3/5 as high as the preceding one. If this ball is dropped from a height of 15 feet, what is the total distance it has traveled after it hits the surface the fifth time?
Dwn.Up.Dwn.Up.Dwn..Up...Dwn...Up...Dwn
15..9...9.5.4.5.4.3.24.3.24.1.944.1.944
Ssumming yields 54.168 = 54 21/125 feet.

Guys, here is the answer.
Ball is dropped from a height of 15 feet and each bounce it goes to 3/5 as high as the preceding one.
Now, total distance traveled in:
1st drop:15 (till it hits the ground)
2nd drop:(15*3/5)=9*2=18 [Multiply by 2 as distance to go up (15*3/5) and to come down (15*3/5),so multiply by 2]
3rd drop: (9*3/5)=5.4*2 =10.8
4th drop: (5.4*3/5)=3.24*2 =6.48
5th drop: (3.24*3/5)=1.94*2= 3.888 (now ball will hit the ground after 5th fall)
total distance: distance in 1st drop+ 2nd drop+3rd drop+4th drop+5th drop.
= 15+18+10.8+6.48+3.888=54.168
Hope this helps.... 
It's a sort og geometric series with limiting sum. Please I need a perfect formula to calculate it instead of the normal addition

each time a bouncy ball bounce, it bounces to half the height from which it had just fallen. if jack drops a ball from a building and it rises to a height of 24 metres, how high will it bounce on the fifth bounce?