SOLUTION: A reboncing ball rebounces each time to a height equal to one half the height of the previous bounce. If it is dropped from a height of 16 mtrs. find the total distance it has trav

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Question 109830: A reboncing ball rebounces each time to a height equal to one half the height of the previous bounce. If it is dropped from a height of 16 mtrs. find the total distance it has travelled when it hits ground for the 10th time.
Found 2 solutions by Edwin McCravy, bucky:
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!

    
From the picture, you can see that the ball falls 10 times 
but it only rises 9 times.

The "falls" form this geometric series with 10 terms, with
first term 16, common ratio 1/2:

16+8+4+2+1+1/2+1/4+1/8+1/16+1/32

The "rises" form this geometric series with 9 terms, with
first term 8, common ratio 1/2

8+4+2+1+1/2+1/4+1/8+1/16+1/32

The only difference in the two is the first "fall" of 16 ft. 
Let's find the sum of the geometric series of "rises" and then 
we can find the sum of the geometric series of "falls" by adding 
the first 16 foot "fall" to it:

The formula for the sum of a geometric series is

Sn = a1(1-rn)/(1-r)

For the "rises", a1 = 8, r = 1/2, n = 9

S9 = 8[1-(1/2)9]/(1-1/2)

S9 = 8[1-1/512]/(1/2)

S9 = 8[511/512]×(2/1)

S9 = (8×511×2)/(512)

S9 = 511/32 feet = sum of "rises" only.

The sum of the "falls" accounts for 16 feet more
than the "rises" or

Sum of "falls" = 511/32 + 16 = 511/32 + 512/32 = 1023/32 feet

So, "sum of rises" + "sum of falls" = 511/32 + 1023/32 = 

1534/32 = 767/16 feet = 47.9375 feet the ball travels.

Edwin
Answer by bucky(2189)   (Show Source): You can put this solution on YOUR website!
You can work this problem at least two ways. One way is to just do a table analysis of
what happens ... as follows
.
Ball falls 16 feet and hits the floor first time. 16 feet ... 1 time
.
Next ball rebounds to 8 ft and falls back 8 ft where it hits floor second time. Total distance
traveled to this point is 16 + 8 + 8 = 32 ft and 2 hits
.
Ball rebounds to 4 ft and falls back 4 ft where it hits 3rd time. Total distance is 32 + 4 + 4
= 40 ft and 3 hits
.
Ball rebounds to 2 ft and falls back 2 ft where it hits 4th time. Total distance now is
40 + 2 + 2 = 44 ft and 4 hits.
.
Ball rebounds to 1 ft and falls back 1 ft where it hits 5th time. Total distance is now
44 + 1 + 1 = 46 ft and 5 hits
.
Ball rebounds to 1/2 ft and falls back 1/2 ft where it hits 6th time. Total distance up
to this point is 46 + 1/2 + 1/2 = 47 ft and 6 hits.
.
Ball rebounds to 1/4 ft and falls back 1/4 ft to hit 7th time. Total distance now is
47 + 1/4 + 1/4 = 47.5 ft and 7 hits
.
Ball rebounds to 1/8 ft (0.125 ft) and falls back 1/8 ft (again 0.125 ft) and hits for the
8th time. Total distance now is 47.5 + 0.125 + 0.125 = 47.75 ft and 8 hits.
.
The ball rebounds to 1/16 ft (0.0625 ft) and falls back the same distance. The total distance
is now 47.75 + 0.0625 + 0.0625 = 47.875 ft and 9 hits.
.
Finally the ball rebounds to 1/32 ft (0.03125 ft) and falls back the same amount, hitting
the floor for the 10th and final time. Total distance = 47.875 + 0.03125 + 0.03125 = 47.9325 ft.
.
You could also do this problem using a geometric progression.
.
Suppose that instead of dropping the ball from a height of 16 ft to start the problem, that
we shot it up 16 feet from floor level. The first cycle would be up 16 ft and down 16 ft for
a total of 32 ft. The next cycle would be up 8 ft and down 8 ft for a total of 16 ft. Each
cycle thereafter would be half of the preceding cycle. So in this progression the first
term is 32 ft, the next 16, the next 8, and so on.
.
If you define a as the first term, r as the common ratio (in this problem r = 1/2), n
as the number of terms to be considered (10 for this problem) and L as the last term, the
equation for the sum (S) of the terms in a geometric progression is:
.

.
in which
.
In this problem:
.

.
Now that we have L, we can return to the sum equation and substitute appropriate values:
.

.
So this tells us that the total distance traveled is 63.9375 ft. But don't forget that
in order to get this into the form of a geometric progression we added 16 feet on the
front end so that every cycle would be complete. Therefore, we need to subtract that 16 feet
off because the ball did not start at ground level. It was dropped from the height of
16 feet. Subtracting 16 feet results in 63.9375 ft - 16 ft = 47.9375 ft for the total
distance traveled during the cycles. This agrees with the answer we got the first way
we did it ...
.
Hope this helps you to understand the problem and how it can be solved.
.
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