Question

A ball is dropped from a height of 192 inches onto a level floor. After the third bounce it is still 3 inches off the ground. Presuming that the height the ball bounces is always the same fraction of the height reached on the previous bounce, what is that fraction?

Answer 1

Answer: The required fraction = $\dfrac18$

Step-by-step explanation:

Let the required fraction = $\dfrac{p}{q}$

Given: Initial height = 192 inches

Height of ball after second bounce = $\dfrac{p}{q}\times192$

Height of ball after third bounce = $\dfrac{p}{q}\times\dfrac{p}{q}\times192=192\dfrac{p^2}{q^2}$

After the third bounce it is 3 inches off the ground.

So,

$(\dfrac{p}{q})^2192=3\\\\\\(\dfrac{p}{q})^2=\dfrac{3}{192}\\\\(\dfrac{p}{q})^2=\dfrac{1}{64}\\\\(\dfrac{p}{q})^2=(\dfrac{1}{8})^2\\\\ \dfrac{p}{q}=\dfrac{1}{8}$

Hence, The required fraction = $\dfrac18$