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When a ball is dropped onto a flat floor, it bounces to 65% of the height from which it was dropped. If the ball is dropped from 80 cm, construct a sequence consists of the first, second and third term which represents the height of each bounce from the first drop.
13n
Show that the height of nth bounce, Tn is = 80  .
20
Hence, find the height of the fifth bounce

Respuesta :

Using a geometric sequence, it is found that:

  • The rule for the height of the nth bounce is:

[tex]a_n = 52(0.65)^{n-1}[/tex]

  • The height of the fifth bounce is of 9.28 cm.

In a geometric sequence, the quotient between consecutive terms is always the same, called common ratio q.

The nth term of a geometric sequence is given by:

[tex]a_n = a_1q^{n-1}[/tex]

In which [tex]a_1[/tex] is the first term.

In this problem:

  • Bounces to 65% of the height from which it was dropped, thus the common ratio is of [tex]q = 0.65[/tex].
  • Dropped from 80 cm, thus, the height of the first bounce is [tex]a_1 = 80(0.65) = 52[/tex].

Thus, the rule for the height of the nth bounce is:

[tex]a_n = 52(0.65)^{n-1}[/tex]

The height of the fifth bounce is [tex]a_5[/tex], thus:

[tex]a_5 = 52(0.65)^{5-1} = 9.28[/tex]

The height of the fifth bounce is of 9.28 cm.

A similar problem is given at https://brainly.com/question/11847927

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