# A ball is dropped from a height of 10 feet and bounces. Each bounce is 3/4 of the height of the bounce before. Thus, after the ball hits the floor for the first time, the ball rises to a height of 10(3/4) = 7.5 feet, and after it hits the second floor for the second time, it rises to a height of 7.5(3/4) = 10(3/4)² = 5.625 feet. (Assume g = 32 ft/sec² and that there is no air resistance.)

1) Find an expression for the height to which the ball rises after it hits the floor for the nth time.

2) Find an expression for the total vertical distance the ball has traveled when it hits the floor for the first, second, third, and fourth times.

3) Find an expression for the total vertical distance the ball has traveled when it hits the floor for the nth time.

**Solution:**

Given, height = 10 feet.

Each bounce is 3/4 of the height of the previous bounce.

1) we have to find an expression for the height to which the ball rises after it hits the floor for the nth time.

Given, for the first time, the ball rises to a height of 10(3/4) = 7.5 feet

For the second time, the ball rises to a height of 7.5(3/4) = 10(3/4)^{2} = 5.625 feet

The general expression is 10, 10(3/4), 10(3/4)^{2},.......,10(3/4)^{n}.

So, \(a_{n}=(10)(\frac{3}{4})^{n}\)

Therefore, an expression for the height to which the ball rises after it hits the floor for the nth time is \(a_{n}=(10)(\frac{3}{4})^{n}\)

2) we have to find an expression for the total vertical distance the ball has traveled when it hits the floor for the first, second, third, and fourth times.

For the first time, the ball hits the floor at 10 feet.

For the second time, the ball hits the floor at the distance = 10 + [10(3/4) + 10(3/4)]

= 10 + [30/4 + 30/4]

= 10 + [60/4]

= 10 + 15

= 25 feet

For the third time, the ball hits the floor at the distance = 10 + [10(3/4) + 10(3/4)] + [10(3/4)^{2} + 10(3/4)^{2}]

= 25 + [90/16 + 90/16]

= 25 + [180/16]

= 25 + 11.25

= 36.25 feet.

For the fourth time, the ball hits the floor at the distance = 10 + [10(3/4) + 10(3/4)] + [10(3/4)^{2} + 10(3/4)^{2}] + [10(3/4)^{3} + 10(3/4)^{3}]

= 36.25 + [10(27/64) + 10(27/64)]

= 36.25 + 4.219 + 4.219

= 44.688 feet

Therefore, expression for the total vertical distance the ball has travelled when it hits the floor for the first, second, third and fourth times is 10 + [10(3/4) + 10(3/4)] + [10(3/4)^{2} + 10(3/4)^{2}] + [10(3/4)^{3} + 10(3/4)^{3}]

3) We have to find an expression for the total vertical distance the ball has traveled when it hits the floor for the nth time.

Sum = 10 + [10(3/4) + 10(3/4)] + [10(3/4)^{2} + 10(3/4)^{2}] + [10(3/4)^{3} + 10(3/4)^{3}] + ….. + [10(3/4)^{n} + 10(3/4)^{n}]

So, sum = 10 + [20(3/4)] + [20(3/4)^{2}] + …… + [20(3/4)^{n}]

Sum = 10 + 20[(3/4) + (3/4)^{2} + (3/4)^{3} + ……. + (3/4)^{n}]

The expression 20[(3/4) + (3/4)^{2} + (3/4)^{3} + ……. + (3/4)^{n}] resembles the sum of geometric series.

\(s_{n}=\frac{a(1-r^{n})}{(1-r)}\) where, r < 1

So, sum = \(10+20[\frac{\frac{3}{4}(1-(\frac{3}{4})^{n})}{(1-\frac{3}{4})}]\)

= \(10+20[\frac{\frac{3}{4}(1-(\frac{3}{4})^{n})}{\frac{1}{4}}]\)

= \(10+20[\frac{3}{4}\times \frac{4}{1}(1-(\frac{3}{4})^{n})]\)

= \(10+20[3(1-(\frac{3}{4})^{n})]\)

Therefore, an expression for the total vertical distance the ball has traveled when it hits the floor for the nth time is \(10+20[3(1-(\frac{3}{4})^{n})]\).

## A ball is dropped from a height of 10 feet and bounces. Each bounce is 3/4 of the height of the bounce before. Thus, after the ball hits the floor for the first time, the ball rises to a height of 10(3/4) = 7.5 feet, and after it hits the second floor for the second time, it rises to a height of 7.5(3/4) = 10(3/4)^{2} = 5.625 feet. (Assume g = 32 ft/sec^{2} and that there is no air resistance.)

**Summary:**

A ball is dropped from a height of 10 feet and bounces. Each bounce is 3/4 of the height of the bounce before. Thus, after the ball hits the floor for the first time, the ball rises to a height of 10(3/4) = 7.5 feet, and after it hits the second floor for the second time, it rises to a height of 7.5(3/4) = 10(3/4)^{2} = 5.625 feet. (Assume g = 32 ft/sec^{2} and that there is no air resistance.)

- An expression for the height to which the ball rises after it hits the floor for the nth time is \(a_{n}=(10)(\frac{3}{4})^{n}\)
- An expression for the total vertical distance the ball has traveled when it hits the floor for the first, second, third, and fourth times is 10 + [10(3/4) + 10(3/4)] + [10(3/4)
^{2}+ 10(3/4)^{2}] + [10(3/4)^{3}+ 10(3/4)^{3}] - An expression for the total vertical distance the ball has traveled when it hits the floor for the nth time is \(10+20[3(1-(\frac{3}{4})^{n})]\).

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