a rubber ball fell from a high of 16 feet and then its rebounded until it rested, such that each time it rebounded to a high =1/2 the high rebounded to previouslly
a) find the high rebounded to after the 10th term
b)find the distance cover by the ball before it rested
To find the height rebounded to after the 10th term and the distance covered by the ball before it rested, we can use geometric sequences.
Let's start by finding the height rebounded to after each term.
Given:
Initial height = 16 feet
Each subsequent rebounded height is half of the previous rebounded height.
a) Finding the height rebounded to after the 10th term:
To find the height rebounded to after the nth term, we can use the formula:
hn = h0 * r^(n-1)
where hn is the height rebounded to after the nth term, h0 is the initial height, r is the common ratio, and n is the number of terms.
In this case:
h0 = 16 feet (initial height)
r = 1/2 (each subsequent rebounded height is half of the previous rebounded height)
n = 10 (10th term)
Plugging these values into the formula:
h10 = 16 * (1/2)^(10-1)
h10 = 16 * (1/2)^9
h10 = 16 * (1/512)
h10 = 16/32
h10 = 1/2 feet
Therefore, the height rebounded to after the 10th term is 1/2 feet.
b) Finding the distance covered by the ball before it rested:
To find the distance covered by the ball before it rested, we can sum all the heights rebounded to until it rests.
The sum of a finite geometric series can be found using the formula:
Sn = a * (1 - r^n) / (1 - r)
where Sn is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms.
In this case:
a = 16 feet (initial height)
r = 1/2 (each subsequent rebounded height is half of the previous rebounded height)
n = number of terms until the ball rested
Since the ball rebounds until it rests, we need to find the value of n.
The height rebounds after the nth term can be considered negligible when it is less than 1/16 feet (which is the height of the smallest unit). Let's find the smallest n that satisfies this condition:
h_n = 16 * (1/2)^(n-1) < 1/16
16/2^(n-1) < 1/16
2^(n-1) > 16 * 16
2^(n-1) > 256
From this, we find n > 9.
Therefore, n = 10 is the smallest value that ensures the height rebounded after the 10th term is larger than 1/16 feet.
Plugging these values into the formula:
S = 16 * (1 - (1/2)^10) / (1 - 1/2)
S = 16 * (1 - 1/1024) / (1/2)
S = 16 * (1023/1024) / (1/2)
S = (16 * 1023) / (1024/2)
S = (16 * 1023) / 512
S = 32 * 1023
Therefore, the distance covered by the ball before it rested is 32 * 1023 feet.
Note: The sum of the series above represents the total distance covered by the ball, including both upward and downward motion.