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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?

Sagot :

ParthKohliid
This is a geometric sequence.  The first term is the max height of the first curved path, which is 0.5.  The second one is 52% of that meaning that it is 0.52 times the first term.  The third term is 0.52 times the second term. Thus, in this geometric sequence,

[tex]a = 0.5 [/tex]
[tex]r = 0.52 [/tex]

You will need to use the relation [tex] a_n = a \cdot r^{n-1} [/tex]
InesWalstonid

Answer:

  1. [tex]f(n)=0.5(0.52)^{n-1}[/tex]
  2. 0.14 m

Step-by-step explanation:

The initial height of the ball is 0.5 m

Each curved path has 52% of the height of the previous path, i.e the height of the ball after one bounce will be,

[tex]=\dfrac{52}{100}\times 0.5\\\\=0.52\times 0.5\ m[/tex]

The height of the ball after 2 bounces will be,

[tex]=\dfrac{52}{100}\times(0.52\times 0.5)[/tex]

[tex]=0.52\times0.52\times 0.5[/tex]

[tex]=0.52^2\times 0.5\ m[/tex]

Hence the series becomes,

[tex]0.5,0.5(0.52),0.5(0.52)^2,............[/tex]

This is the case of Geometric Progression.

But as it is given that the initial height will be given by n=1, so the rules for finding the height f(n) after n bounces would be,

[tex]f(n)=0.5(0.52)^{n-1}[/tex]

Putting n=3, we can get the height of the ball of the third path,

[tex]\Rightarrow f(3)=0.5(0.52)^{3-1}=0.5(0.52)^{2}=0.14\ m[/tex]

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