You drop a ball from a height of 0.5 meter. Each curved path has 52% of the height of the previous path. a. Write a rule for the sequence using centimeters. The initial height is given by the term n = 1. b. What height will the ball be at the top of the third path?

The initial height is .5 m. If the ball only reaches 52% of the previous max height, then after the third bounce you have:
.5 x (.52)^3=0.070304 meters as the height of the ball
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lauradevilleros lauradevilleros · Answer 2 · 2019-08-02T03:24:44+02:00

Answer:

A rule for the sequence using centimeters is [tex]h_{n}=0.52^{n-1}*h_{1} \\[/tex]

and the height will the ball be at the top of the third path is 13.52 cm

Step-by-step explanation:

You know that the initial height (h) is given by the term n = 1, then:

[tex]h_{1}=0.5m=50cm[/tex]

[tex]h_{2}=50cm*0.52[/tex]

52%=52/100=0.52

[tex]h_{3}=50cm*0.52*0.52=50cm*0.52^2[/tex]

[tex]h_{4}=50cm*0.52*0.52*0.52=50cm*0.52^3[/tex]

[tex]h_{5}=50cm*0.52*0,52*0.52*0.52=50cm*0.52^4[/tex]

Then the standard equation is:

[tex]h_{n}=0.52^{n-1}*h_{1} \\[/tex]

Now you need to calculate [tex]h_{3}[/tex]

[tex]h_{3}=0.52^{3-1}*h_{1}=0.52^2*(50cm)=13.52 cm[/tex]