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nordsb [41]
3 years ago
7

3

Mathematics
1 answer:
emmainna [20.7K]3 years ago
6 0

Answer:

B

Step-by-step explanation:

cause on T4L you have to round it to the nearest whole and I put B and it was correct.

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2-7 is not = 0a

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A rubber ball dropped on a hard surface takes a sequence of bounces, each one 1/6 as high as the preceding one. If this ball is
Brums [2.3K]

Answer:

\frac{259}{54}\text{ or }4.8\text{feet}

Step-by-step explanation:

GIVEN: A rubber ball dropped on a hard surface takes a sequence of bounces, each one \frac{1}{6} as high as the preceding one.

TO FIND: If this ball is dropped from a height of 12 feet, how far will it have traveled when it hits the surface the fifth time.

SOLUTION:

once the ball is dropped on hard surface it bounces \frac{1}{6} of preceding one and comes down the same distance.

When the ball is dropped from 12 feet height

after first hit =12\times\frac{1}{6}\text{ up}+12\times\frac{1}{6}\text{ down}=4

new height =12\times\frac{1}{6}=2\text{feet}

after second hit =2\times\frac{1}{6}\text{ up}+2\times\frac{1}{6}\text{ down}=\frac{2}{3}

new height =2\times\frac{1}{6}=\frac{1}{3}\text{ feet}

after third hit  =\frac{1}{3}\times\frac{1}{6}\text{ up}+\frac{1}{3}\times\frac{1}{6}\text{ down}=\frac{1}{9}

new height =\frac{1}{3}\times\frac{1}{6}=\frac{1}{18}\text{ feet}

after fourth hit =\frac{1}{18}\times\frac{1}{6}\text{ up}+\frac{1}{18}\times\frac{1}{6}\text{ down}=\frac{1}{54}

adding all distance =4+\frac{2}{3}+\frac{1}{9}+\frac{1}{54}

                                 =\frac{259}{54} feet

Hence the ball will travel   \frac{259}{54} feet before it hits the surface fifth time.

                                 

4 0
3 years ago
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