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A ball fell vertically downward from a height of 270 centimeters above the ground.
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Each time the ball touches the ground, it bounces four-ninths of the previous distance.
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Find the falling distance when the ball touches the ground at the fourth time.
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Then find the total distance the ball covered during its motion.
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Give all answers to the nearest unit.
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We are told that a ball falls downward from a height of 270 centimeters above the ground.
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It then bounces four-ninths of the previous distance.
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Four-ninths of 270 centimeters is 120 centimeters.
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This means that the ball bounces back to a height of 120 centimeters above the ground before falling back to the ground.
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The height that the ball falls from can be represented as a sequence where π sub one, the first term, is 270.
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The second term, π sub two, is equal to 120, which we obtained by multiplying 270 by four-ninths.
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We can calculate the third term by multiplying 120 by four-ninths.
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This is equal to 160 over three or 53.33 recurring.
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The first part of our question asks us to find the falling distance when the ball touches the ground at the fourth time.
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This is the value of π sub four, so we can simply multiply π sub three by four-ninths.
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This is equal to 23.70 and so on.
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As we are asked to give our answer to the nearest unit, we need to round up.
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The falling distance when the ball touches the ground at the fourth time to the nearest unit is 24 centimeters.
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An alternative method to calculate this would be to recognize we have a geometric sequence.
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This is any sequence that has a common ratio or multiplier between consecutive terms.
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In this question, the common ratio is equal to four-ninths.
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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.
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In this question, we are trying to calculate the fourth term.
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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.
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The second part of our question asks us to find the total distance the ball covered.
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There are a few ways of calculating this.
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However, we will start by calculating the total distance the ball traveled downward.
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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.
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This is equal to the first term π sub one divided by one minus π.
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Substituting in our values, we have 270 divided by one minus four-ninths.
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This is equal to 270 divided by five-ninths, which is 486.
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The ball fell a total downward distance of 486 centimeters during its motion.
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We might initially think this is the final answer.
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However, we also need to consider the distance traveled when the ball rebounded back upwards.
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After the ball hit the ground for the first time, it rebounded 120 centimeters upwards.
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After it hit the ground a second time, it rebounded 53.3 recurring centimeters upwards.
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This continues in the same geometric sequence as before.
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In the ballβs motion upwards, we have another geometric sequence, this time with first term 120.
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The common difference is still four-ninths.
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So the total distance traveled in an upward direction is 120 divided by one minus four-ninths.
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This is equal to 216.
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The total distance that the ball traveled in an upward direction was 216 centimeters.
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We can therefore calculate the total distance by adding 486 and 216.
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This is equal to 702 centimeters.
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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.
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486 multiplied by two minus 270 is also equal to 702.
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The falling distance the ball traveled when it touched the ground the fourth time was 24 centimeters.
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And the total distance the ball covered during its entire motion was 702 centimeters.