A rubber ball is dropped from a certain height. It is found to rebounce only $90 %$ of its previous height. If it is dropped from the top of a $25 m$ tall building, to what height would it raise after bounding on the ground two times.

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Solution

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
$A = P {(1 + \frac{R}{100})}^{n}$
$A = 25 {(1 - \frac{10}{100})}^{2}$
$= 25 {(\frac{90}{100})}^{2}$
$= 20.25 m$

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

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=2The height to which it raises after bouncing two times on the groundA=P(1+R100)nA=25(1−10100)2=25(90100)2=20.25m

A rubber ball is dropped from a certain height. It is found to rebounce only $90 %$ of its previous height. If it is dropped from the top of a $25 m$ tall building, to what height would it raise after bounding on the ground two times.