a rubber ball is dropped froma building that is 16 meters high. Each time the ball bounces, it bounces up half as high as the previous bounce. It is caught by Romer, the wonder dog, when it bounces 1 meter high. How many meters did the ball travel before it was caught?
height: 16;8;4;2;1
fourth bounce. Distance went 16+2*8+2*4+2*2+1
or another way:
Distance=16+ 16(.5)+16*.5^2+ 16*.5^3+1
this looks like a geometric series, r=.5
So in a geometric series,
an= A1*r^n n=0, 1, 2,..
to find our problem
1=16 (.5)^n
1/16= (1/2)^n
take log base 1/2 of each side..
4=n
To find out how many meters the ball traveled before it was caught, we need to add up the distances it traveled during each bounce.
Since the ball is dropped from a height of 16 meters, the first bounce will reach a height of 16 meters.
Then, for each subsequent bounce, the ball reaches a height that is half of the previous bounce. So the heights for each bounce can be represented as a geometric sequence: 16, 8, 4, 2, 1 (where the last bounce is 1 meter high).
To find the total distance traveled, we can calculate the sum of this geometric sequence.
The formula to find the sum of a geometric sequence is given by: S = a * (1 - r^n) / (1 - r),
Where:
S = the sum of the geometric sequence,
a = the first term of the sequence,
r = the common ratio of the sequence,
n = the number of terms in the sequence.
In this case, the first term (a) is 16, the common ratio (r) is 0.5 since each bounce is half as high as the previous bounce, and we have a total of 5 terms (n).
Plugging these values into the formula, we get:
S = 16 * (1 - 0.5^5) / (1 - 0.5)
Simplifying, we get:
S = 16 * (1 - 0.03125) / 0.5
S = 16 * 0.96875 / 0.5
S = 15.5 meters
Therefore, the rubber ball traveled a total distance of 15.5 meters before it was caught by Romer, the wonder dog.