Respuesta :
Answer:
[tex]\frac{259}{54}\text{ or }4.8\text{feet}[/tex]
Step-by-step explanation:
GIVEN: A rubber ball dropped on a hard surface takes a sequence of bounces, each one [tex]\frac{1}{6}[/tex] as high as the preceding one.
TO FIND: If this ball is dropped from a height of [tex]12[/tex] feet, how far will it have traveled when it hits the surface the fifth time.
SOLUTION:
once the ball is dropped on hard surface it bounces [tex]\frac{1}{6}[/tex] of preceding one and comes down the same distance.
When the ball is dropped from [tex]12[/tex] feet height
after first hit [tex]=12\times\frac{1}{6}\text{ up}+12\times\frac{1}{6}\text{ down}=4[/tex]
new height [tex]=12\times\frac{1}{6}=2\text{feet}[/tex]
after second hit [tex]=2\times\frac{1}{6}\text{ up}+2\times\frac{1}{6}\text{ down}=\frac{2}{3}[/tex]
new height [tex]=2\times\frac{1}{6}=\frac{1}{3}\text{ feet}[/tex]
after third hit [tex]=\frac{1}{3}\times\frac{1}{6}\text{ up}+\frac{1}{3}\times\frac{1}{6}\text{ down}=\frac{1}{9}[/tex]
new height [tex]=\frac{1}{3}\times\frac{1}{6}=\frac{1}{18}\text{ feet}[/tex]
after fourth hit [tex]=\frac{1}{18}\times\frac{1}{6}\text{ up}+\frac{1}{18}\times\frac{1}{6}\text{ down}=\frac{1}{54}[/tex]
adding all distance [tex]=4+\frac{2}{3}+\frac{1}{9}+\frac{1}{54}[/tex]
[tex]=\frac{259}{54}[/tex] feet
Hence the ball will travel [tex]\frac{259}{54}[/tex] feet before it hits the surface fifth time.
