Question

A rubber ball is dropped vertically on the ground from a certain height. It is found to rise only 90% of its previous height. If it is dropped from the top of a 25 m tall building,to what height would it rise after bouncing in the ground two times?

Answer 1

Answer:

Step-by-step explanation:

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

Step-by-step explanation:

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

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

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

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

$$= 20.25 m$$