Question

lled by the ball until it stops bouncing a ball bounces from a height of 2 metres and returns to 80% of its previous height on each bounce. find the total distance travelled by the ball until it stops bouncing

Answer 1

 The solution to the problem is as follows:

We have 2+.8(2) + .8(.8(2)) + .8(.8(.8(2))) + ... = 

2( .8^0 + .8^1 + .8^2 + .8^3 + ... ) = 

2(.8^n -1) / (.8-1) . As n-->infinity, .8^n-->0 giving us 

2(-1)/(-.2) = 2(5) = 10 meters.


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Answer 2

Answer:

The distance traveled by the ball until it stops bouncing is:

                                   18 meters

Step-by-step explanation:

It is given that:

a ball bounces from a height of 2 meters and returns to 80% of its previous height on each bounce.

This means that the distance traveled by the ball is the distance it travels by going down as well as coming up on bouncing.

and is given as follows:

$\text{Total Distance}=2+0.8\times 2\ up+0.8\times 2\ down+(0.8)^2\times 2\ up+(0.8)^2\times 2\ down+.....\\\\\text{Total distance}=2+4\times (0.8)+4\times (0.8)^2+4\times (0.8)^3+....\\\\\text{Total distance}=2+4\times [0.8+(0.8)^2+(0.8)^3+......]\\\\\text{Total distance}=2+4\times (\dfrac{0.8}{1-0.8})$

Since, the formula of infinite geometric progression is:

$\sum_{n=1}^{\infty} ar^{n-1}=\dfrac{a}{1-r}$

i.e.

$\text{Total distance}=2+4\times (\dfrac{0.8}{0.2})\\\\\text{Total distance}=2+4\times 4\\\\\text{Total distance}=2+16\\\\\text{Total distance}=18\ \text{meters}$