A ball is dropped from a height of 5 ft. Assuming that on each bounce the ball rebounds to one-third of its previous height, find the total distance travelled by the ball.
Only one bounce? If so, D = 5 + 2(1/3 of 5) = ? If more bounces, extrapolate.
wala kang tulog
To find the total distance traveled by the ball, we need to calculate the sum of the distances covered during each bounce.
Given that the ball rebounds to one-third of its previous height, we can set up a geometric sequence to represent the distances covered during each bounce.
The first term of the sequence is the initial height from which the ball is dropped, which is 5 ft. The common ratio is one-third (1/3), as the ball rebounds to one-third of its previous height.
Let's calculate the distances covered during each bounce:
First bounce: 5 ft
Second bounce: (1/3) * 5 ft = (5/3) ft
Third bounce: (1/3) * (5/3) ft = (5/9) ft
Fourth bounce: (1/3) * (5/9) ft = (5/27) ft
We can observe that the distances covered during each bounce form a geometric sequence with the first term (a) as 5 ft and the common ratio (r) as 1/3.
Using the formula for the sum of an infinite geometric series, we can calculate the total distance traveled by the ball.
The formula for the sum of an infinite geometric series is given by:
S = a / (1 - r)
where S is the sum of the series, a is the first term, and r is the common ratio.
In this case, a = 5 ft and r = 1/3.
Let's plug in the values and calculate the sum:
S = 5 ft / (1 - 1/3)
S = 5 ft / (2/3)
S = 5 ft * (3/2)
S = 15/2 ft
Therefore, the total distance traveled by the ball is 15/2 ft or 7.5 ft.