Answer:
More than 6.21 times
Step-by-step explanation:
The common ratio [tex]r=\dfrac{4}{5}[/tex]
Distance of first bounce
[tex]a_1=4\times \dfrac{4}{5}=\dfrac{16}{5}[/tex]
Second bounce
[tex]a_2=ar^{n-1}=\dfrac{16}{5}(\dfrac{4}{5})^{2-1}\\\Rightarrow a_2=\dfrac{16}{5}(\dfrac{4}{5})[/tex]
n bounce
[tex]a_n=ar^{n-1}\\\Rightarrow a_n=\dfrac{16}{5}(\dfrac{4}{5})^{n-1}[/tex]
Now the nth bounce will be less than 1 feet
[tex]\dfrac{16}{5}(\dfrac{4}{5})^{n-1}<1\\\Rightarrow (\dfrac{4}{5})^{n-1}<\dfrac{5}{16}[/tex]
Applying logarithms on both sides we get
[tex](n-1)\ln \dfrac{4}{5}<\ln\dfrac{5}{16}\\\Rightarrow n>\dfrac{\ln\dfrac{5}{16}}{\ln \dfrac{4}{5}}+1[/tex]
The inequality changes as both [tex]\ln\dfrac{5}{16}[/tex] and [tex]\ln \dfrac{4}{5}[/tex] are negative numbers.
So,
[tex]n>6.21[/tex]
Hence, the number of bounces is more than 6.21 times.
