Question 15, Exercise 4.5

Solutions of Question 15 of Exercise 4.5 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

To test its elasticity, a rubber ball is dropped into a $30 ft$ hollow tube that is calibrated so that the scientist can measure the height of each subsequent bounce. The scientist found that on each bounce, the ball rises to a height $\frac{2}{5}$ the height of the previous bounce. How far will the ball travel before it stops bouncing?

Solution.

Hight of ball $= 30 ft$
First rebound $= 30 \times \frac{2}{5} = 12 ft$
Second rebound $= 12 \times \frac{2}{5} = \frac{24}{5} ft$
Third rebound $= \frac{24}{5} \times \frac{2}{5} = \frac{48}{25} ft$

Let $D$ be the total distance covered by the ball. Then $$D=30+2\left(12+\frac{24}{5}+\frac{24}{5}+... \right)$$ To find the sum of infinite geometric series $$ 12+\frac{24}{5}+\frac{24}{5}+... $$ We have $a_1=12$, $r=\frac{2}{5}$ with $|r|<1$, thus \begin{align*} S_\infty & = \frac{a_1}{1-r} \\ & = \frac{12}{1-\frac{2}{5}} = \frac{60}{3} & = 20. \end{align*} Hence $$D=30+2(20) = 70$$ Hence ball will travel $70\,\, ft$ before it stops bouncing.

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