Question

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

Answer 1

Using a geometric sequence, it is found that:

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

$a_n = 52(0.65)^{n-1}$

• 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:

$a_n = a_1q^{n-1}$

In which $a_1$ is the first term.

In this problem:

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

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

$a_n = 52(0.65)^{n-1}$

The height of the fifth bounce is $a_5$, thus:

$a_5 = 52(0.65)^{5-1} = 9.28$

The height of the fifth bounce is of 9.28 cm.

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