Answer:
1.125 m
Step-by-step explanation:
Given:
Initial height of the ball, [tex]h_{0}=72\textrm{ m}[/tex]
Height after first bounce, [tex]h_{1}=36\textrm{ m}[/tex]
Height after second bounce, [tex]h_{2}=18\textrm{ m}[/tex]
So, the reduction in height after each bounce is half of the previous one. Thus, it follows a geometric sequence with the first term as 72 and common ratio of [tex]r=\frac{1}{2}[/tex].
Therefore, the nth term of a geometric sequence is given as:
[tex]h_{n}=h_{0}r^{n}[/tex]
Here, [tex]n=6,r=\frac{1}{2},h_{0}=72[/tex]. This gives,
[tex]h_{6}=72\times (\frac{1}{2})^{6}\\h_{6}=\frac{72}{64}=1.125\textrm{ m}[/tex].
Therefore, the height of the ball after sixth bounce is 1.125 m.
