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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 :

JeanaShupp

Answer: The required fraction = [tex]\dfrac18[/tex]

Step-by-step explanation:

Let the required fraction = [tex]\dfrac{p}{q}[/tex]

Given: Initial height = 192 inches

Height of ball after second bounce = [tex]\dfrac{p}{q}\times192[/tex]

Height of ball after third bounce = [tex]\dfrac{p}{q}\times\dfrac{p}{q}\times192=192\dfrac{p^2}{q^2}[/tex]

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

So,

[tex](\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}[/tex]

Hence, The required fraction = [tex]\dfrac18[/tex]

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