A ball is dropped from 16 feet above a hard surface. After each time it hits the surface, it rebounds to a height that is 3/4 of its previous maximum height. What is the total vertical distance, to the nearest foot, the ball has traveled when it strikes the ground for the 10th time? Write out the first five terms of this sum to help visualize.

The rebound of the ball from the height is an illustration of a geometric series

(a) The total vertical distance

The given parameters are:

-- the first term

-- the common ratio

-- number of terms

The total vertical distance when it strikes for the 10th time is the sum of nth terms of the geometric series.

So, we have:

Substitute values for (a), (r) and n

Simplify

Hence, the total vertical distance at the 10th rebound is 60.352 feet

(b) The first five terms of this sum

The nth term of a geometric series is:

So, we have:

Hence, the first five terms of the sum are 16, 12, 9, 6.75 and 5.0625

Read more about geometric series at:

link

a) Total distance travelled to the nearest foot = 105ft

b) First 5 terms = 12, 9, 6.75, 5.0625, 3.796875

Step-by-step explanation:

From the question above, we can see that this is a geometric series or progression

Formula for geometric progression:

Tn = arⁿ-¹

Where

a = first term

r = common ration

n = number of terms

From the question, we are told that the ball dropped from 16 feet

Hence First term(a) = 16

We are told that the ball rebounds to 3/4 of it's previous height anytime it hits the surface.

Hence, common ratio(r) = 3/4

a) We are told to find total distance to the nearest foot that the ball travelled when it strikes the ground for the 10th time

This means we are to find the sum of the first 10 terms.

Because the ball rises and falls, we would use the rises form

Where the first term = 12 and the number of times is 9 because it rises 9 times

The formula we would use to find this sum is based on if the common ratio is less than or greater than 1

Our common ratio = 3/4, hence it is less than 1.

When r< 1, sum of n terms =

Sn = a(1 - rⁿ)/ 1 - r

It is important to

Where a = 12

r = 3/4

n = 9

Sn = 12( 1 - (3/4)^9) / 1- 3/4

Sn = 104.79ft

Approximately = 105ft

b) First five terms

Using the formula

Tn = arⁿ-¹

Where

a = first term = 12

r = common ration

n = number of terms

First term, n = 1 = 12 × 3/4^(1-1)

= 12 × 3/4 ^(0)

= 12× 1

= 12

Second term, n = 2 = 12 × 3/4^(2-1)

= 12 × 3/4 ^(1)

= 36/4

= 9

Third term, n = 3 = 12 × 3/4^(3-1)

= 12 × (3/4)²

= 6.75

Fourth term, n = 4= 12 × 3/4^(4-1)

= 12 × (3/4)³

= 5.0625

Fifth term, n = 5 = 12 × 3/4^(5-1)

= 12 × (3/4)⁴

= 3.796875

Therefore, the first 5 terms = 12, 9, 6.75, 5.0625, 3.796875

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