Question

A rubber ball is dropped from a height of 10 feet onto a hard surface. After the first bounce, the ball is 6 feet in the air and continues to follow a sequence of bounces. Assume the distances from the ground form a geometric sequence. What is the total DOWNWARD vertical distance the ball has traveled after it hits the surface the 5th time?

Answer 1

Answer:

The total DOWNWARD vertical distance the ball has traveled after it hits the surface the 5th time is 23.056 feet.

Step-by-step explanation:

Geometric sequence:

The nth term is a product of the previous(nth-1) term and the common ratio. That is:

$a_n = qa_{n-1}$

In which q is the common difference.

Sum of the first n terms of geometric sequence:

The sum of the first n terms of a geometric sequence is given by:

$S_n = \frac{a_1(1-q^n)}{1-q}$

A rubber ball is dropped from a height of 10 feet onto a hard surface.

This means that $a_1 = 10$

The ball is 6 feet in the air and continues to follow a sequence of bounces

This means that $q = \frac{6}{10} = 0.6$

What is the total DOWNWARD vertical distance the ball has traveled after it hits the surface the 5th time?

This is $S_5$. So

$S_5 = \frac{10(1-0.6^5)}{1-0.6} = 23.056$

The total DOWNWARD vertical distance the ball has traveled after it hits the surface the 5th time is 23.056 feet.