Practice · Bouncing ball rises a fraction of its fall · Sensim Math

Variant 1 easy

A ball bounces back up to $\frac{1}{3}$ of the height it falls from. If this ball is dropped from a height of $90\ \text{m}$ , what is the total distance the ball travels up to the moment it bounces up for the second time, in meters?

Your answer

Show solution

1 · Understandwhat's really being asked

A ball is dropped from 90 m. Each bounce sends it back up to $\frac{1}{3}$ of the height it just fell. I need the total distance the ball travels (down plus up) from the start until the instant it reaches the top of its second upward bounce.

Givens

The ball rebounds to $\frac{1}{3}$ of the height it falls from.
It is dropped from a height of 90 m.
We follow the motion until it bounces up for the second time.

Unknowns

The total distance traveled (in meters) up to the moment of the second upward bounce.

Constraints

Distance is the sum of every fall and every rise, all counted as positive lengths.
Up to the second bounce-up the path is: first fall, first rise, second fall, second rise.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Sketching the up-and-down path makes clear exactly which legs to add. Each rebound height is a fraction-of-a-number subproblem (divide by the denominator, multiply by the numerator), and the answer is the sum of the four legs.

3 · Execute5 carry out the plan

1First fall

#1 Draw a Diagram 3.NF.A.1

The ball drops the full starting height of 90 m.

90\ \text{m}

The first leg is simply the height it is released from.

2First rise (first bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{3}

of 90 m. Find

\frac{1}{3}

of 90 by dividing by 3, then multiply by 1.

90 \div 3 = 30,\quad 30 \times 1 = 30

A fraction of a number means split into equal parts, then take that many parts.

3Second fall

#1 Draw a Diagram 3.OA.A.2

The ball falls back down from its first bounce height, so it drops the same 30 m it rose.

30\ \text{m}

Whatever height it reaches, it falls that same distance straight back down.

4Second rise (second bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{3}

of the 30 m it just fell. Divide 30 by 3, then multiply by 1.

30 \div 3 = 10,\quad 10 \times 1 = 10

Same fraction rule applied to the new, smaller fall height.

5Add all four legs

#7 Identify Subproblems 3.OA.A.2

Total distance is the first fall, first rise, second fall, and second rise added together.

90 + 30 + 30 + 10 = 160

Total path length is just every up-and-down leg summed.

Answer: 160 m

4 · Reviewdoes it hold up?

Each bounce is smaller than the last (90, 30, 10), which fits a ball losing height. The total 160 m is sensible since the ball goes down 90, then makes two shorter up-down trips. Units stay in meters throughout.

Another way: Look for a pattern (tool 5): rebound heights shrink by the factor $\frac{1}{3}$ each time (90, 30, 10...). Listing fall and rise legs up to the second bounce and summing gives the same 160 m.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Finding $\frac{1}{3}$ of each fall height by splitting into 3 equal parts and taking 1 of them.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing heights by 3 and adding the four legs of the path.

💡Takeaway. This only needs Grade 3 fraction sense: split by 3, take 1 parts, and add up each up-and-down trip!

Variant 2 easy

A ball bounces back up to $\frac{1}{5}$ of the height it falls from. If this ball is dropped from a height of $100\ \text{m}$ , what is the total distance the ball travels up to the moment it bounces up for the second time, in meters?

Your answer

Show solution

1 · Understandwhat's really being asked

A ball is dropped from 100 m. Each bounce sends it back up to $\frac{1}{5}$ of the height it just fell. I need the total distance the ball travels (down plus up) from the start until the instant it reaches the top of its second upward bounce.

Givens

The ball rebounds to $\frac{1}{5}$ of the height it falls from.
It is dropped from a height of 100 m.
We follow the motion until it bounces up for the second time.

Unknowns

The total distance traveled (in meters) up to the moment of the second upward bounce.

Constraints

Distance is the sum of every fall and every rise, all counted as positive lengths.
Up to the second bounce-up the path is: first fall, first rise, second fall, second rise.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Sketching the up-and-down path makes clear exactly which legs to add. Each rebound height is a fraction-of-a-number subproblem (divide by the denominator, multiply by the numerator), and the answer is the sum of the four legs.

3 · Execute5 carry out the plan

1First fall

#1 Draw a Diagram 3.NF.A.1

The ball drops the full starting height of 100 m.

100\ \text{m}

The first leg is simply the height it is released from.

2First rise (first bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{5}

of 100 m. Find

\frac{1}{5}

of 100 by dividing by 5, then multiply by 1.

100 \div 5 = 20,\quad 20 \times 1 = 20

A fraction of a number means split into equal parts, then take that many parts.

3Second fall

#1 Draw a Diagram 3.OA.A.2

The ball falls back down from its first bounce height, so it drops the same 20 m it rose.

20\ \text{m}

Whatever height it reaches, it falls that same distance straight back down.

4Second rise (second bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{5}

of the 20 m it just fell. Divide 20 by 5, then multiply by 1.

20 \div 5 = 4,\quad 4 \times 1 = 4

Same fraction rule applied to the new, smaller fall height.

5Add all four legs

#7 Identify Subproblems 3.OA.A.2

Total distance is the first fall, first rise, second fall, and second rise added together.

100 + 20 + 20 + 4 = 144

Total path length is just every up-and-down leg summed.

Answer: 144 m

4 · Reviewdoes it hold up?

Each bounce is smaller than the last (100, 20, 4), which fits a ball losing height. The total 144 m is sensible since the ball goes down 100, then makes two shorter up-down trips. Units stay in meters throughout.

Another way: Look for a pattern (tool 5): rebound heights shrink by the factor $\frac{1}{5}$ each time (100, 20, 4...). Listing fall and rise legs up to the second bounce and summing gives the same 144 m.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Finding $\frac{1}{5}$ of each fall height by splitting into 5 equal parts and taking 1 of them.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing heights by 5 and adding the four legs of the path.

💡Takeaway. This only needs Grade 3 fraction sense: split by 5, take 1 parts, and add up each up-and-down trip!

Variant 3 easy

A ball bounces back up to $\frac{1}{5}$ of the height it falls from. If this ball is dropped from a height of $125\ \text{m}$ , what is the total distance the ball travels up to the moment it bounces up for the second time, in meters?

Your answer

Show solution

1 · Understandwhat's really being asked

A ball is dropped from 125 m. Each bounce sends it back up to $\frac{1}{5}$ of the height it just fell. I need the total distance the ball travels (down plus up) from the start until the instant it reaches the top of its second upward bounce.

Givens

The ball rebounds to $\frac{1}{5}$ of the height it falls from.
It is dropped from a height of 125 m.
We follow the motion until it bounces up for the second time.

Unknowns

The total distance traveled (in meters) up to the moment of the second upward bounce.

Constraints

Distance is the sum of every fall and every rise, all counted as positive lengths.
Up to the second bounce-up the path is: first fall, first rise, second fall, second rise.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Sketching the up-and-down path makes clear exactly which legs to add. Each rebound height is a fraction-of-a-number subproblem (divide by the denominator, multiply by the numerator), and the answer is the sum of the four legs.

3 · Execute5 carry out the plan

1First fall

#1 Draw a Diagram 3.NF.A.1

The ball drops the full starting height of 125 m.

125\ \text{m}

The first leg is simply the height it is released from.

2First rise (first bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{5}

of 125 m. Find

\frac{1}{5}

of 125 by dividing by 5, then multiply by 1.

125 \div 5 = 25,\quad 25 \times 1 = 25

A fraction of a number means split into equal parts, then take that many parts.

3Second fall

#1 Draw a Diagram 3.OA.A.2

The ball falls back down from its first bounce height, so it drops the same 25 m it rose.

25\ \text{m}

Whatever height it reaches, it falls that same distance straight back down.

4Second rise (second bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{5}

of the 25 m it just fell. Divide 25 by 5, then multiply by 1.

25 \div 5 = 5,\quad 5 \times 1 = 5

Same fraction rule applied to the new, smaller fall height.

5Add all four legs

#7 Identify Subproblems 3.OA.A.2

Total distance is the first fall, first rise, second fall, and second rise added together.

125 + 25 + 25 + 5 = 180

Total path length is just every up-and-down leg summed.

Answer: 180 m

4 · Reviewdoes it hold up?

Each bounce is smaller than the last (125, 25, 5), which fits a ball losing height. The total 180 m is sensible since the ball goes down 125, then makes two shorter up-down trips. Units stay in meters throughout.

Another way: Look for a pattern (tool 5): rebound heights shrink by the factor $\frac{1}{5}$ each time (125, 25, 5...). Listing fall and rise legs up to the second bounce and summing gives the same 180 m.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Finding $\frac{1}{5}$ of each fall height by splitting into 5 equal parts and taking 1 of them.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing heights by 5 and adding the four legs of the path.

💡Takeaway. This only needs Grade 3 fraction sense: split by 5, take 1 parts, and add up each up-and-down trip!

Variant 4 easy

A ball bounces back up to $\frac{2}{5}$ of the height it falls from. If this ball is dropped from a height of $150\ \text{m}$ , what is the total distance the ball travels up to the moment it bounces up for the second time, in meters?

Your answer

Show solution

1 · Understandwhat's really being asked

A ball is dropped from 150 m. Each bounce sends it back up to $\frac{2}{5}$ of the height it just fell. I need the total distance the ball travels (down plus up) from the start until the instant it reaches the top of its second upward bounce.

Givens

The ball rebounds to $\frac{2}{5}$ of the height it falls from.
It is dropped from a height of 150 m.
We follow the motion until it bounces up for the second time.

Unknowns

The total distance traveled (in meters) up to the moment of the second upward bounce.

Constraints

Distance is the sum of every fall and every rise, all counted as positive lengths.
Up to the second bounce-up the path is: first fall, first rise, second fall, second rise.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Sketching the up-and-down path makes clear exactly which legs to add. Each rebound height is a fraction-of-a-number subproblem (divide by the denominator, multiply by the numerator), and the answer is the sum of the four legs.

3 · Execute5 carry out the plan

1First fall

#1 Draw a Diagram 3.NF.A.1

The ball drops the full starting height of 150 m.

150\ \text{m}

The first leg is simply the height it is released from.

2First rise (first bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{2}{5}

of 150 m. Find

\frac{1}{5}

of 150 by dividing by 5, then multiply by 2.

150 \div 5 = 30,\quad 30 \times 2 = 60

A fraction of a number means split into equal parts, then take that many parts.

3Second fall

#1 Draw a Diagram 3.OA.A.2

The ball falls back down from its first bounce height, so it drops the same 60 m it rose.

60\ \text{m}

Whatever height it reaches, it falls that same distance straight back down.

4Second rise (second bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{2}{5}

of the 60 m it just fell. Divide 60 by 5, then multiply by 2.

60 \div 5 = 12,\quad 12 \times 2 = 24

Same fraction rule applied to the new, smaller fall height.

5Add all four legs

#7 Identify Subproblems 3.OA.A.2

Total distance is the first fall, first rise, second fall, and second rise added together.

150 + 60 + 60 + 24 = 294

Total path length is just every up-and-down leg summed.

Answer: 294 m

4 · Reviewdoes it hold up?

Each bounce is smaller than the last (150, 60, 24), which fits a ball losing height. The total 294 m is sensible since the ball goes down 150, then makes two shorter up-down trips. Units stay in meters throughout.

Another way: Look for a pattern (tool 5): rebound heights shrink by the factor $\frac{2}{5}$ each time (150, 60, 24...). Listing fall and rise legs up to the second bounce and summing gives the same 294 m.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Finding $\frac{2}{5}$ of each fall height by splitting into 5 equal parts and taking 2 of them.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing heights by 5 and adding the four legs of the path.

💡Takeaway. This only needs Grade 3 fraction sense: split by 5, take 2 parts, and add up each up-and-down trip!

Variant 5 medium

A ball bounces back up to $\frac{1}{4}$ of the height it falls from. If this ball is dropped from a height of $160\ \text{m}$ , what is the total distance the ball travels up to the moment it bounces up for the second time, in meters?

Your answer

Show solution

1 · Understandwhat's really being asked

A ball is dropped from 160 m. Each bounce sends it back up to $\frac{1}{4}$ of the height it just fell. I need the total distance the ball travels (down plus up) from the start until the instant it reaches the top of its second upward bounce.

Givens

The ball rebounds to $\frac{1}{4}$ of the height it falls from.
It is dropped from a height of 160 m.
We follow the motion until it bounces up for the second time.

Unknowns

The total distance traveled (in meters) up to the moment of the second upward bounce.

Constraints

Distance is the sum of every fall and every rise, all counted as positive lengths.
Up to the second bounce-up the path is: first fall, first rise, second fall, second rise.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Sketching the up-and-down path makes clear exactly which legs to add. Each rebound height is a fraction-of-a-number subproblem (divide by the denominator, multiply by the numerator), and the answer is the sum of the four legs.

3 · Execute5 carry out the plan

1First fall

#1 Draw a Diagram 3.NF.A.1

The ball drops the full starting height of 160 m.

160\ \text{m}

The first leg is simply the height it is released from.

2First rise (first bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{4}

of 160 m. Find

\frac{1}{4}

of 160 by dividing by 4, then multiply by 1.

160 \div 4 = 40,\quad 40 \times 1 = 40

A fraction of a number means split into equal parts, then take that many parts.

3Second fall

#1 Draw a Diagram 3.OA.A.2

The ball falls back down from its first bounce height, so it drops the same 40 m it rose.

40\ \text{m}

Whatever height it reaches, it falls that same distance straight back down.

4Second rise (second bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{4}

of the 40 m it just fell. Divide 40 by 4, then multiply by 1.

40 \div 4 = 10,\quad 10 \times 1 = 10

Same fraction rule applied to the new, smaller fall height.

5Add all four legs

#7 Identify Subproblems 3.OA.A.2

Total distance is the first fall, first rise, second fall, and second rise added together.

160 + 40 + 40 + 10 = 250

Total path length is just every up-and-down leg summed.

Answer: 250 m

4 · Reviewdoes it hold up?

Each bounce is smaller than the last (160, 40, 10), which fits a ball losing height. The total 250 m is sensible since the ball goes down 160, then makes two shorter up-down trips. Units stay in meters throughout.

Another way: Look for a pattern (tool 5): rebound heights shrink by the factor $\frac{1}{4}$ each time (160, 40, 10...). Listing fall and rise legs up to the second bounce and summing gives the same 250 m.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Finding $\frac{1}{4}$ of each fall height by splitting into 4 equal parts and taking 1 of them.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing heights by 4 and adding the four legs of the path.

💡Takeaway. This only needs Grade 3 fraction sense: split by 4, take 1 parts, and add up each up-and-down trip!

Variant 6 medium

A ball bounces back up to $\frac{2}{5}$ of the height it falls from. If this ball is dropped from a height of $175\ \text{m}$ , what is the total distance the ball travels up to the moment it bounces up for the second time, in meters?

Your answer

Show solution

1 · Understandwhat's really being asked

A ball is dropped from 175 m. Each bounce sends it back up to $\frac{2}{5}$ of the height it just fell. I need the total distance the ball travels (down plus up) from the start until the instant it reaches the top of its second upward bounce.

Givens

The ball rebounds to $\frac{2}{5}$ of the height it falls from.
It is dropped from a height of 175 m.
We follow the motion until it bounces up for the second time.

Unknowns

The total distance traveled (in meters) up to the moment of the second upward bounce.

Constraints

Distance is the sum of every fall and every rise, all counted as positive lengths.
Up to the second bounce-up the path is: first fall, first rise, second fall, second rise.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Sketching the up-and-down path makes clear exactly which legs to add. Each rebound height is a fraction-of-a-number subproblem (divide by the denominator, multiply by the numerator), and the answer is the sum of the four legs.

3 · Execute5 carry out the plan

1First fall

#1 Draw a Diagram 3.NF.A.1

The ball drops the full starting height of 175 m.

175\ \text{m}

The first leg is simply the height it is released from.

2First rise (first bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{2}{5}

of 175 m. Find

\frac{1}{5}

of 175 by dividing by 5, then multiply by 2.

175 \div 5 = 35,\quad 35 \times 2 = 70

A fraction of a number means split into equal parts, then take that many parts.

3Second fall

#1 Draw a Diagram 3.OA.A.2

The ball falls back down from its first bounce height, so it drops the same 70 m it rose.

70\ \text{m}

Whatever height it reaches, it falls that same distance straight back down.

4Second rise (second bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{2}{5}

of the 70 m it just fell. Divide 70 by 5, then multiply by 2.

70 \div 5 = 14,\quad 14 \times 2 = 28

Same fraction rule applied to the new, smaller fall height.

5Add all four legs

#7 Identify Subproblems 3.OA.A.2

Total distance is the first fall, first rise, second fall, and second rise added together.

175 + 70 + 70 + 28 = 343

Total path length is just every up-and-down leg summed.

Answer: 343 m

4 · Reviewdoes it hold up?

Each bounce is smaller than the last (175, 70, 28), which fits a ball losing height. The total 343 m is sensible since the ball goes down 175, then makes two shorter up-down trips. Units stay in meters throughout.

Another way: Look for a pattern (tool 5): rebound heights shrink by the factor $\frac{2}{5}$ each time (175, 70, 28...). Listing fall and rise legs up to the second bounce and summing gives the same 343 m.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Finding $\frac{2}{5}$ of each fall height by splitting into 5 equal parts and taking 2 of them.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing heights by 5 and adding the four legs of the path.

💡Takeaway. This only needs Grade 3 fraction sense: split by 5, take 2 parts, and add up each up-and-down trip!

Variant 7 medium

A ball bounces back up to $\frac{1}{3}$ of the height it falls from. If this ball is dropped from a height of $180\ \text{m}$ , what is the total distance the ball travels up to the moment it bounces up for the second time, in meters?

Your answer

Show solution

1 · Understandwhat's really being asked

A ball is dropped from 180 m. Each bounce sends it back up to $\frac{1}{3}$ of the height it just fell. I need the total distance the ball travels (down plus up) from the start until the instant it reaches the top of its second upward bounce.

Givens

The ball rebounds to $\frac{1}{3}$ of the height it falls from.
It is dropped from a height of 180 m.
We follow the motion until it bounces up for the second time.

Unknowns

The total distance traveled (in meters) up to the moment of the second upward bounce.

Constraints

Distance is the sum of every fall and every rise, all counted as positive lengths.
Up to the second bounce-up the path is: first fall, first rise, second fall, second rise.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Sketching the up-and-down path makes clear exactly which legs to add. Each rebound height is a fraction-of-a-number subproblem (divide by the denominator, multiply by the numerator), and the answer is the sum of the four legs.

3 · Execute5 carry out the plan

1First fall

#1 Draw a Diagram 3.NF.A.1

The ball drops the full starting height of 180 m.

180\ \text{m}

The first leg is simply the height it is released from.

2First rise (first bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{3}

of 180 m. Find

\frac{1}{3}

of 180 by dividing by 3, then multiply by 1.

180 \div 3 = 60,\quad 60 \times 1 = 60

A fraction of a number means split into equal parts, then take that many parts.

3Second fall

#1 Draw a Diagram 3.OA.A.2

The ball falls back down from its first bounce height, so it drops the same 60 m it rose.

60\ \text{m}

Whatever height it reaches, it falls that same distance straight back down.

4Second rise (second bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{3}

of the 60 m it just fell. Divide 60 by 3, then multiply by 1.

60 \div 3 = 20,\quad 20 \times 1 = 20

Same fraction rule applied to the new, smaller fall height.

5Add all four legs

#7 Identify Subproblems 3.OA.A.2

Total distance is the first fall, first rise, second fall, and second rise added together.

180 + 60 + 60 + 20 = 320

Total path length is just every up-and-down leg summed.

Answer: 320 m

4 · Reviewdoes it hold up?

Each bounce is smaller than the last (180, 60, 20), which fits a ball losing height. The total 320 m is sensible since the ball goes down 180, then makes two shorter up-down trips. Units stay in meters throughout.

Another way: Look for a pattern (tool 5): rebound heights shrink by the factor $\frac{1}{3}$ each time (180, 60, 20...). Listing fall and rise legs up to the second bounce and summing gives the same 320 m.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Finding $\frac{1}{3}$ of each fall height by splitting into 3 equal parts and taking 1 of them.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing heights by 3 and adding the four legs of the path.

💡Takeaway. This only needs Grade 3 fraction sense: split by 3, take 1 parts, and add up each up-and-down trip!

Variant 8 medium

A ball bounces back up to $\frac{2}{5}$ of the height it falls from. If this ball is dropped from a height of $200\ \text{m}$ , what is the total distance the ball travels up to the moment it bounces up for the second time, in meters?

Your answer

Show solution

1 · Understandwhat's really being asked

A ball is dropped from 200 m. Each bounce sends it back up to $\frac{2}{5}$ of the height it just fell. I need the total distance the ball travels (down plus up) from the start until the instant it reaches the top of its second upward bounce.

Givens

The ball rebounds to $\frac{2}{5}$ of the height it falls from.
It is dropped from a height of 200 m.
We follow the motion until it bounces up for the second time.

Unknowns

The total distance traveled (in meters) up to the moment of the second upward bounce.

Constraints

Distance is the sum of every fall and every rise, all counted as positive lengths.
Up to the second bounce-up the path is: first fall, first rise, second fall, second rise.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Sketching the up-and-down path makes clear exactly which legs to add. Each rebound height is a fraction-of-a-number subproblem (divide by the denominator, multiply by the numerator), and the answer is the sum of the four legs.

3 · Execute5 carry out the plan

1First fall

#1 Draw a Diagram 3.NF.A.1

The ball drops the full starting height of 200 m.

200\ \text{m}

The first leg is simply the height it is released from.

2First rise (first bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{2}{5}

of 200 m. Find

\frac{1}{5}

of 200 by dividing by 5, then multiply by 2.

200 \div 5 = 40,\quad 40 \times 2 = 80

A fraction of a number means split into equal parts, then take that many parts.

3Second fall

#1 Draw a Diagram 3.OA.A.2

The ball falls back down from its first bounce height, so it drops the same 80 m it rose.

80\ \text{m}

Whatever height it reaches, it falls that same distance straight back down.

4Second rise (second bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{2}{5}

of the 80 m it just fell. Divide 80 by 5, then multiply by 2.

80 \div 5 = 16,\quad 16 \times 2 = 32

Same fraction rule applied to the new, smaller fall height.

5Add all four legs

#7 Identify Subproblems 3.OA.A.2

Total distance is the first fall, first rise, second fall, and second rise added together.

200 + 80 + 80 + 32 = 392

Total path length is just every up-and-down leg summed.

Answer: 392 m

4 · Reviewdoes it hold up?

Each bounce is smaller than the last (200, 80, 32), which fits a ball losing height. The total 392 m is sensible since the ball goes down 200, then makes two shorter up-down trips. Units stay in meters throughout.

Another way: Look for a pattern (tool 5): rebound heights shrink by the factor $\frac{2}{5}$ each time (200, 80, 32...). Listing fall and rise legs up to the second bounce and summing gives the same 392 m.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Finding $\frac{2}{5}$ of each fall height by splitting into 5 equal parts and taking 2 of them.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing heights by 5 and adding the four legs of the path.

💡Takeaway. This only needs Grade 3 fraction sense: split by 5, take 2 parts, and add up each up-and-down trip!

Variant 9 hard

A ball bounces back up to $\frac{1}{5}$ of the height it falls from. If this ball is dropped from a height of $250\ \text{m}$ , what is the total distance the ball travels up to the moment it bounces up for the second time, in meters?

Your answer

Show solution

1 · Understandwhat's really being asked

A ball is dropped from 250 m. Each bounce sends it back up to $\frac{1}{5}$ of the height it just fell. I need the total distance the ball travels (down plus up) from the start until the instant it reaches the top of its second upward bounce.

Givens

The ball rebounds to $\frac{1}{5}$ of the height it falls from.
It is dropped from a height of 250 m.
We follow the motion until it bounces up for the second time.

Unknowns

The total distance traveled (in meters) up to the moment of the second upward bounce.

Constraints

Distance is the sum of every fall and every rise, all counted as positive lengths.
Up to the second bounce-up the path is: first fall, first rise, second fall, second rise.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Sketching the up-and-down path makes clear exactly which legs to add. Each rebound height is a fraction-of-a-number subproblem (divide by the denominator, multiply by the numerator), and the answer is the sum of the four legs.

3 · Execute5 carry out the plan

1First fall

#1 Draw a Diagram 3.NF.A.1

The ball drops the full starting height of 250 m.

250\ \text{m}

The first leg is simply the height it is released from.

2First rise (first bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{5}

of 250 m. Find

\frac{1}{5}

of 250 by dividing by 5, then multiply by 1.

250 \div 5 = 50,\quad 50 \times 1 = 50

A fraction of a number means split into equal parts, then take that many parts.

3Second fall

#1 Draw a Diagram 3.OA.A.2

The ball falls back down from its first bounce height, so it drops the same 50 m it rose.

50\ \text{m}

Whatever height it reaches, it falls that same distance straight back down.

4Second rise (second bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{5}

of the 50 m it just fell. Divide 50 by 5, then multiply by 1.

50 \div 5 = 10,\quad 10 \times 1 = 10

Same fraction rule applied to the new, smaller fall height.

5Add all four legs

#7 Identify Subproblems 3.OA.A.2

Total distance is the first fall, first rise, second fall, and second rise added together.

250 + 50 + 50 + 10 = 360

Total path length is just every up-and-down leg summed.

Answer: 360 m

4 · Reviewdoes it hold up?

Each bounce is smaller than the last (250, 50, 10), which fits a ball losing height. The total 360 m is sensible since the ball goes down 250, then makes two shorter up-down trips. Units stay in meters throughout.

Another way: Look for a pattern (tool 5): rebound heights shrink by the factor $\frac{1}{5}$ each time (250, 50, 10...). Listing fall and rise legs up to the second bounce and summing gives the same 360 m.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Finding $\frac{1}{5}$ of each fall height by splitting into 5 equal parts and taking 1 of them.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing heights by 5 and adding the four legs of the path.

💡Takeaway. This only needs Grade 3 fraction sense: split by 5, take 1 parts, and add up each up-and-down trip!

Variant 10 hard

A ball bounces back up to $\frac{1}{3}$ of the height it falls from. If this ball is dropped from a height of $270\ \text{m}$ , what is the total distance the ball travels up to the moment it bounces up for the second time, in meters?

Your answer

Show solution

1 · Understandwhat's really being asked

A ball is dropped from 270 m. Each bounce sends it back up to $\frac{1}{3}$ of the height it just fell. I need the total distance the ball travels (down plus up) from the start until the instant it reaches the top of its second upward bounce.

Givens

The ball rebounds to $\frac{1}{3}$ of the height it falls from.
It is dropped from a height of 270 m.
We follow the motion until it bounces up for the second time.

Unknowns

The total distance traveled (in meters) up to the moment of the second upward bounce.

Constraints

Distance is the sum of every fall and every rise, all counted as positive lengths.
Up to the second bounce-up the path is: first fall, first rise, second fall, second rise.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Sketching the up-and-down path makes clear exactly which legs to add. Each rebound height is a fraction-of-a-number subproblem (divide by the denominator, multiply by the numerator), and the answer is the sum of the four legs.

3 · Execute5 carry out the plan

1First fall

#1 Draw a Diagram 3.NF.A.1

The ball drops the full starting height of 270 m.

270\ \text{m}

The first leg is simply the height it is released from.

2First rise (first bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{3}

of 270 m. Find

\frac{1}{3}

of 270 by dividing by 3, then multiply by 1.

270 \div 3 = 90,\quad 90 \times 1 = 90

A fraction of a number means split into equal parts, then take that many parts.

3Second fall

#1 Draw a Diagram 3.OA.A.2

The ball falls back down from its first bounce height, so it drops the same 90 m it rose.

90\ \text{m}

Whatever height it reaches, it falls that same distance straight back down.

4Second rise (second bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{3}

of the 90 m it just fell. Divide 90 by 3, then multiply by 1.

90 \div 3 = 30,\quad 30 \times 1 = 30

Same fraction rule applied to the new, smaller fall height.

5Add all four legs

#7 Identify Subproblems 3.OA.A.2

Total distance is the first fall, first rise, second fall, and second rise added together.

270 + 90 + 90 + 30 = 480

Total path length is just every up-and-down leg summed.

Answer: 480 m

4 · Reviewdoes it hold up?

Each bounce is smaller than the last (270, 90, 30), which fits a ball losing height. The total 480 m is sensible since the ball goes down 270, then makes two shorter up-down trips. Units stay in meters throughout.

Another way: Look for a pattern (tool 5): rebound heights shrink by the factor $\frac{1}{3}$ each time (270, 90, 30...). Listing fall and rise legs up to the second bounce and summing gives the same 480 m.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Finding $\frac{1}{3}$ of each fall height by splitting into 3 equal parts and taking 1 of them.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing heights by 3 and adding the four legs of the path.

💡Takeaway. This only needs Grade 3 fraction sense: split by 3, take 1 parts, and add up each up-and-down trip!

Variant 11 hard

A ball bounces back up to $\frac{1}{4}$ of the height it falls from. If this ball is dropped from a height of $320\ \text{m}$ , what is the total distance the ball travels up to the moment it bounces up for the second time, in meters?

Your answer

Show solution

1 · Understandwhat's really being asked

A ball is dropped from 320 m. Each bounce sends it back up to $\frac{1}{4}$ of the height it just fell. I need the total distance the ball travels (down plus up) from the start until the instant it reaches the top of its second upward bounce.

Givens

The ball rebounds to $\frac{1}{4}$ of the height it falls from.
It is dropped from a height of 320 m.
We follow the motion until it bounces up for the second time.

Unknowns

The total distance traveled (in meters) up to the moment of the second upward bounce.

Constraints

Distance is the sum of every fall and every rise, all counted as positive lengths.
Up to the second bounce-up the path is: first fall, first rise, second fall, second rise.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Sketching the up-and-down path makes clear exactly which legs to add. Each rebound height is a fraction-of-a-number subproblem (divide by the denominator, multiply by the numerator), and the answer is the sum of the four legs.

3 · Execute5 carry out the plan

1First fall

#1 Draw a Diagram 3.NF.A.1

The ball drops the full starting height of 320 m.

320\ \text{m}

The first leg is simply the height it is released from.

2First rise (first bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{4}

of 320 m. Find

\frac{1}{4}

of 320 by dividing by 4, then multiply by 1.

320 \div 4 = 80,\quad 80 \times 1 = 80

A fraction of a number means split into equal parts, then take that many parts.

3Second fall

#1 Draw a Diagram 3.OA.A.2

The ball falls back down from its first bounce height, so it drops the same 80 m it rose.

80\ \text{m}

Whatever height it reaches, it falls that same distance straight back down.

4Second rise (second bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{4}

of the 80 m it just fell. Divide 80 by 4, then multiply by 1.

80 \div 4 = 20,\quad 20 \times 1 = 20

Same fraction rule applied to the new, smaller fall height.

5Add all four legs

#7 Identify Subproblems 3.OA.A.2

Total distance is the first fall, first rise, second fall, and second rise added together.

320 + 80 + 80 + 20 = 500

Total path length is just every up-and-down leg summed.

Answer: 500 m

4 · Reviewdoes it hold up?

Each bounce is smaller than the last (320, 80, 20), which fits a ball losing height. The total 500 m is sensible since the ball goes down 320, then makes two shorter up-down trips. Units stay in meters throughout.

Another way: Look for a pattern (tool 5): rebound heights shrink by the factor $\frac{1}{4}$ each time (320, 80, 20...). Listing fall and rise legs up to the second bounce and summing gives the same 500 m.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Finding $\frac{1}{4}$ of each fall height by splitting into 4 equal parts and taking 1 of them.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing heights by 4 and adding the four legs of the path.

💡Takeaway. This only needs Grade 3 fraction sense: split by 4, take 1 parts, and add up each up-and-down trip!

Variant 12 hard

A ball bounces back up to $\frac{1}{4}$ of the height it falls from. If this ball is dropped from a height of $480\ \text{m}$ , what is the total distance the ball travels up to the moment it bounces up for the second time, in meters?

Your answer

Show solution

1 · Understandwhat's really being asked

A ball is dropped from 480 m. Each bounce sends it back up to $\frac{1}{4}$ of the height it just fell. I need the total distance the ball travels (down plus up) from the start until the instant it reaches the top of its second upward bounce.

Givens

The ball rebounds to $\frac{1}{4}$ of the height it falls from.
It is dropped from a height of 480 m.
We follow the motion until it bounces up for the second time.

Unknowns

The total distance traveled (in meters) up to the moment of the second upward bounce.

Constraints

Distance is the sum of every fall and every rise, all counted as positive lengths.
Up to the second bounce-up the path is: first fall, first rise, second fall, second rise.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Sketching the up-and-down path makes clear exactly which legs to add. Each rebound height is a fraction-of-a-number subproblem (divide by the denominator, multiply by the numerator), and the answer is the sum of the four legs.

3 · Execute5 carry out the plan

1First fall

#1 Draw a Diagram 3.NF.A.1

The ball drops the full starting height of 480 m.

480\ \text{m}

The first leg is simply the height it is released from.

2First rise (first bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{4}

of 480 m. Find

\frac{1}{4}

of 480 by dividing by 4, then multiply by 1.

480 \div 4 = 120,\quad 120 \times 1 = 120

A fraction of a number means split into equal parts, then take that many parts.

3Second fall

#1 Draw a Diagram 3.OA.A.2

The ball falls back down from its first bounce height, so it drops the same 120 m it rose.

120\ \text{m}

Whatever height it reaches, it falls that same distance straight back down.

4Second rise (second bounce up)

#7 Identify Subproblems 3.NF.A.1

It rebounds to

\frac{1}{4}

of the 120 m it just fell. Divide 120 by 4, then multiply by 1.

120 \div 4 = 30,\quad 30 \times 1 = 30

Same fraction rule applied to the new, smaller fall height.

5Add all four legs

#7 Identify Subproblems 3.OA.A.2

Total distance is the first fall, first rise, second fall, and second rise added together.

480 + 120 + 120 + 30 = 750

Total path length is just every up-and-down leg summed.

Answer: 750 m

4 · Reviewdoes it hold up?

Each bounce is smaller than the last (480, 120, 30), which fits a ball losing height. The total 750 m is sensible since the ball goes down 480, then makes two shorter up-down trips. Units stay in meters throughout.

Another way: Look for a pattern (tool 5): rebound heights shrink by the factor $\frac{1}{4}$ each time (480, 120, 30...). Listing fall and rise legs up to the second bounce and summing gives the same 750 m.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Finding $\frac{1}{4}$ of each fall height by splitting into 4 equal parts and taking 1 of them.
3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing heights by 4 and adding the four legs of the path.

💡Takeaway. This only needs Grade 3 fraction sense: split by 4, take 1 parts, and add up each up-and-down trip!

Bouncing ball rises a fraction of its fall · 12 practice problems

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