Question

The distance travelled in (m) by a ball dropped from a height are 128/9,32/3,8,6,...
How much distance will it travel before coming to rest

Answer 1

Answer: it will trave 56.89 meters before coming to rest.

Step-by-step explanation:

This is a geometric progression since the distance travelled (height) by the ball is reducing by a constant ratio, r. Since the number of times that the ball will bounce is infinite, then we would apply the formula for determining the sum of the terms in a geometric progression to infinity which is expressed as

S = a/(1 - r)

where

S = sum of the distance travelled by the ball

a = initial distance or height of the ball

r = common ratio

From the information given,

a = 128/9

r = (32/3)/(128/9) = 0.75

Therefore,

S = (128/9)/(1 - 0.75) = 56.89 meters

Answer 2

The distance travelled is an illustration of the sum to infinity of a geometric sequence.

The ball will a travel 56.89 meters before coming to rest

The sequence is given as:

$\mathbf{128/9, 32/3, 8, 6....}$

From the sequence above, we have:

$\mathbf{a = 128/9}$ --- the first term

$\mathbf{r = 6/8 = 3/4}$ -- the common ratio

The sum to infinity of a geometric progression is:

$\mathbf{S_{\infty} = \frac{a}{1-r}}$

So, we have:

$\mathbf{S_{\infty} = \frac{128/9}{1-3/4}}$

$\mathbf{S_{\infty} = \frac{128/9}{1/4}}$

Divide

$\mathbf{S_{\infty} = \frac{128}{9} \times 4}$

$\mathbf{S_{\infty} = \frac{128\times 4}{9} }$

$\mathbf{S_{\infty} = \frac{512}{9} }$

$\mathbf{S_{\infty} = 56.89}$

Hence, the ball will a travel 56.89 meters before coming to rest

Read more about sum to infinity at:

https://brainly.com/question/18522826