Question

A ball is dropped from a height of 3 ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen. (a) Find the total distance the ball has traveled at the instant it hits the ground the fifth time. (Enter an exact number.) (b) Find a formula for the total distance the ball has traveled at the instant it hits the ground the nth time.

Answer 1

The total distance the ball has traveled at the instant it hits the ground the fifth time is 4.537 ft. The formula for the total distance the ball has traveled at the instant it hits the ground the nth time is 9 * ((1 - 1/3^n) / (2/3)).

Step-by-step explanation:

Let's solve this problem step by step:

Part (a):

The total distance the ball has traveled at the instant it hits the ground the fifth time is the sum of the distances it has traveled during each bounce. Since the ball always bounces up one-third the distance it has fallen, the series of distances it has traveled can be represented as follows:

1. First bounce: 3 ft
2. Second bounce: (1/3) * 3 ft = 1 ft
3. Third bounce: (1/3) * 1 ft = 1/3 ft
4. Fourth bounce: (1/3) * (1/3) ft = 1/9 ft
5. Fifth bounce: (1/3) * (1/9) ft = 1/27 ft

To find the total distance, we sum up these distances:

Total distance = 3 ft + 1 ft + 1/3 ft + 1/9 ft + 1/27 ft

Simplifying the expression, we get:

Total distance = 3 + 1 + 1/3 + 1/9 + 1/27 ft = 4.537 ft

Part (b):

To find a formula for the total distance the ball has traveled when it hits the ground the nth time, we can set up a geometric series.

Let D_n be the distance traveled during the nth bounce. Since the ball always bounces up one-third the distance it has fallen, we have:

D_n = (1/3) * D_(n-1) where D_1 = 3 ft.

The total distance traveled up to the nth bounce can be represented as:

Total distance = D_1 + D_2 + D_3 + ... + D_n.

To find a formula for the total distance, we can sum up the geometric series:

Total distance = D_1 + D_2 + D_3 + ... + D_n = (1/3) * D_1 + (1/3) * (1/3) * D_1 + (1/3) * (1/3) * (1/3) * D_1 + ... + (1/3)^(n-1) * D_1.

The formula for the sum of a geometric series is:

Sum = a * (1 - r^n) / (1 - r) where a is the first term and r is the common ratio.

Substituting the values, we have:

Total distance = 3 * (1 - (1/3)^n) / (1 - 1/3)

Simplifying the expression, we get:

Total distance = 9 * ((1 - 1/3^n) / (2/3))

Therefore, the formula for the total distance the ball has traveled at the instant it hits the ground the nth time is 9 * ((1 - 1/3^n) / (2/3)).