Question

a rubber ball is dropped on a hard surface takes a sequence of bounces, each one 3/5 as high as the preceeding one. If this ball is dropped at 10 feet, what is the total vertical distance it has traveled at the time when it hits the surface for the fifth time?

Answer 1

Answer:

StepThe answer is 36 14/125 feet

-by-step explanation:

Answer 2

The total vertical distance it has traveled at the time when rubber ball  hits the surface for the fifth time = 34.7 feet

A rubber ball is dropped on a hard surface  

Each one 3/5 as high as the preceding one.  

The initial height dropped = 10 feet

$\rm Height\; achieved \; in \; the \; second \; drop = 10\times (3/5) \\\\ Height\; achieved \; in \; the \; third \; drop = 10\times (3/5)^2 \\\\\rm Height\; achieved \; in \; the \; fourth \; drop = 10\times (3/5)^3\\\\\rm Height\; achieved \; in \; the \; fifth \; drop = 10\times (3/5)^4$

So the total vertical  distance traveled

In the second ,third and fourth tip it travels the vertical  distance two times hence we can write

$10 + 2\times 10 (3/5) + 2\times 10 (3/5)^2 + 2\times10 (3/5)^3 + 10 \times (3/5)^4$

= 10 + 12 + 7.2 + 4.2 + 1.296 = 34.696 feet $\rm \bold \approx {34.7 \; feet}$

hence

The total vertical distance it has traveled at the time when rubber ball  hits the surface for the fifth time = 34.7 feet

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https://brainly.com/question/13008517