contestada

A ball is dropped from a height of 10 feet and returns to a height that is one-half of the height from which it fell. The ball continues to bounce half the height of the previous bounce each time. How far will the ball have traveled when it hits the ground for the fourth time?

Respuesta :

apologiabiology
ok, so this is a sum of aruthmetic sequence and also use some logic

it goes 10 feet down, then 5 up, 5 down, etc (see attachment)

so the total distance is the sum of all the numbers in the sequence doubled, minus the first term (since it didn't bounce 10 feet up, it started from top)

the sum of an aruthmetic is
Sn=[tex] \frac{a_{1}(1-r^n)}{1-r} [/tex]
where a1=first term, r=common ratio and n=which time

first term=10
common ratio is 0.5
n=4

so
2[[tex] \frac{10(1-0.5^4)}{1-0.5} [/tex]]-10=answer27.5

the answer is 27.5 feet
Ver imagen apologiabiology

A sequence of values which have subsequent terms given by multiplying the previous term by a constant value is a geometric sequence

  • The ball will travel 1.25 feet when it hits the ground the fourth time

Reason:

The given parameters are;

Height from which the ball is dropped = 10 feet

Height to which the returns = Half of the initial height

Height to which the ball bounces each time = Half the previous height

The height to which the ball travels the fourth time = Required

Solution:

The height the ball travels is given by a geometric progression

First term of the progression, a = 10

Common ratio of the terms, r = [tex]\dfrac{1}{2}[/tex]

The value of the nth term of a geometric progression is given as follows;

  • aₙ = a·r⁽���⁻¹⁾

Therefore, we have;

  • [tex]a_4 =10 \times \left( \dfrac{1}{2} \right)^{(4 - 1)} = 10 \times \left( \dfrac{1}{2} \right)^{3} = \dfrac{10}{8} = 1.25[/tex]

The height to which the ball will travel when it hits the ground the fourth time, a₄ = 1.25 feet

Learn more here:

https://brainly.com/question/16350673

Otras preguntas