How far does the golf ball travel before coming to rest?

A golf ball is dropped from a height of 81 inches. It rebounds to 2/3 of its original height and continues rebounding in this manner. How far does it travel before coming to rest?|||sv ,





EDIT: (and Rocky, too)...





have given you a wonderful solution to your geometric series question. The other answers are so much "hand waving" junk. Those folks have no idea what is going on in the problem...forget their prattle!!





There are many examples of infinite processes that have finite results. This is one example.





The old joke: A man closes the distance between himself and his beautiful wife by 1/2 each time he moves. The ignorant say "He never reaches her." He answers "I get close enough."





-Fred|||thoereticaly speaking it never come to a rest (assuming it bounces exatly 2/3 its last bounce every time, scince you can divide any number (no matter how small).)|||it travelled = 81[1 + 2{2/3 + (2/3)^2 + ...}] in


= 81[1 + 2{2/3 / (1 --2/3)}] in


= 405 inches|||depends what you call rest, this is how it goes


1st bounce = 2/3rds of 81 = 54


2nd bounce = 2/3rds of 54 =36


3rd bounce = 2/3rds of 36 = 24


4th ...blah blah blah = 16


5th = 10


6th = 6.66


7th 4.4





and it carries on for a long time|||nice question .


when the golf ball is first dropped it will cover 81 m


then it rebounds to cover 2/3 of 81 =54m


then it falls back 54m


then it rebounds 2/3 of 54 =36m


falls back 36m


so total distance = 81 +(54+54+36+36+.....+0 )


81 +2(an G.P with first term = 54 comm diff = 2/3)


81 +2(162) (USING INFINITE G.P FORMULA )


=405m








hope it helps!best of luck!