Before it rebounds n times we have the inequality
36 (3/4)^n < 1
(3/4)^n < 1/36
n ln 3/4 < ln 1/36
n < ln 1/36 / ln 3/4
n < 12.46
so n = 12
answer is 12 bounces
Answer:
Ball will make 13 bounces before it rebounds less than 1 foot.
Step-by-step explanation:
A ball is dropped from a height of 36 feet.
In each bounce the ball reaches a height that is three quarters of the previous height.
Sequence formed will be 36, 27, 20.25.........
This sequence has a common ratio of [tex]\frac{3}{4}[/tex]
Therefore, sequence will be a geometric sequence.
Explicit formula of this sequence will be
[tex]T_{n}=a(r)^{n}[/tex]
where a = first term of the sequence
r = common ratio
n = number of term
For this sequence formula will be
[tex]T_{n}=36\times (\frac{3}{4})^{n}[/tex]
If this term is less than 1
[tex]36\times (\frac{3}{4})^{n}<1[/tex]
Taking log on both the sides
[tex]log[36\times (\frac{3}{4})^{n}]<log1[/tex]
[tex]log36+nlog(\frac{3}{4} )<log1[/tex]
1.5563 + n(-0.1249) < 0
0.1249n > 1.5563
n > [tex]\frac{1.5563}{0.1249}[/tex]
n > 12.46
n ≈ 13
Therefore, ball will make 13 bounces before it rebounds less than 1 foot.
