Question

a rubber ball is dropped from a certain height it is found to rebound only 90% office previous height if it is dropped from the top of a 25 m tall building to what height would it ride rise after bouncing on the ground two times

Answer 1

The ball rises to a height of $$90\%$$ at the first bounce. So at each bounce, the loss in height is $$10\%$$

So taking $$R = -10\%$$ the problem can be solved.

$$P = 25$$ m and $$n = 2$$

The height to which it raises after bouncing two times on the ground

$$A = P \left ( 1 + \dfrac{R}{100} \right )^n$$

$$A = 25 \left( 1 - \dfrac{10}{100} \right )^2$$

$$= 25 \left ( \dfrac{90}{100} \right )^2$$

$$= 20.25 m$$

Answer 2

Answer:

Step-by-step explanation:

1st rebound height = 27/3=9

2nd rebound height=9/3=3

3rd rebound height=3/3=1

4th rebound height=1/3

5th rebound height=1/9

6th rebound height=1/27

7th rebound height=1/81

8th rebound height=1/243

distance traveled=27+ 2(1/243+1/81+1/27+1/9+1/3+1+3+9)=54m