### Video Transcript

A ball fell vertically downward from a height of 270 centimeters above the ground. Each time the ball touches the ground, it bounces four-ninths of the previous distance. Find the falling distance when the ball touches the ground at the fourth time. Then find the total distance the ball covered during its motion. Give all answers to the nearest unit.

We are told that a ball falls downward from a height of 270 centimeters above the ground. It then bounces four-ninths of the previous distance. Four-ninths of 270 centimeters is 120 centimeters. This means that the ball bounces back to a height of 120 centimeters above the ground before falling back to the ground.

The height that the ball falls from can be represented as a sequence where π sub one, the first term, is 270. The second term, π sub two, is equal to 120, which we obtained by multiplying 270 by four-ninths. We can calculate the third term by multiplying 120 by four-ninths. This is equal to 160 over three or 53.33 recurring.

The first part of our question asks us to find the falling distance when the ball touches the ground at the fourth time. This is the value of π sub four, so we can simply multiply π sub three by four-ninths. This is equal to 23.70 and so on. As we are asked to give our answer to the nearest unit, we need to round up. The falling distance when the ball touches the ground at the fourth time to the nearest unit is 24 centimeters.

An alternative method to calculate this would be to recognize we have a geometric sequence. This is any sequence that has a common ratio or multiplier between consecutive terms. In this question, the common ratio is equal to four-ninths. We know that the general or πth term of a geometric sequence, denoted π sub π, is equal to π sub one multiplied by π to the power of π minus one.

In this question, we are trying to calculate the fourth term. Substituting our values of π sub one and π, this is equal to 270 multiplied by four-ninths cubed, which once again gives us an answer of 23.70 and so on, which rounded to the nearest unit is 24 centimeters.

The second part of our question asks us to find the total distance the ball covered. There are a few ways of calculating this. However, we will start by calculating the total distance the ball traveled downward. Since the absolute value of the common ratio is less than one, we can do this by calculating the sum to β of our sequence. This is equal to the first term π sub one divided by one minus π.

Substituting in our values, we have 270 divided by one minus four-ninths. This is equal to 270 divided by five-ninths, which is 486. The ball fell a total downward distance of 486 centimeters during its motion.

We might initially think this is the final answer. However, we also need to consider the distance traveled when the ball rebounded back upwards. After the ball hit the ground for the first time, it rebounded 120 centimeters upwards. After it hit the ground a second time, it rebounded 53.3 recurring centimeters upwards. This continues in the same geometric sequence as before.

In the ballβs motion upwards, we have another geometric sequence, this time with first term 120. The common difference is still four-ninths. So the total distance traveled in an upward direction is 120 divided by one minus four-ninths. This is equal to 216. The total distance that the ball traveled in an upward direction was 216 centimeters.

We can therefore calculate the total distance by adding 486 and 216. This is equal to 702 centimeters. An alternative method here would have been to have multiplied 486 centimeters by two and then subtracting 270 centimeters, as the ball only traveled this distance in the downward direction. 486 multiplied by two minus 270 is also equal to 702.

The falling distance the ball traveled when it touched the ground the fourth time was 24 centimeters. And the total distance the ball covered during its entire motion was 702 centimeters.