Bouncing ball rises a fraction of its fall

3.NF.A.13.OA.A.2 · take · grade 3

Archetype: Track a Quantity Through Changes · step in a 7-type progression

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?

Show solution

Understand

A ball is dropped from 150 m. Each bounce sends it back up to 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 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.

Plan

#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 5, multiply by 2), and the answer is the sum of the four legs.

Execute

#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.

#7 Identify Subproblems 3.NF.A.1

It rebounds to 2/5 of 150 m. Find 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.

#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.

#7 Identify Subproblems 3.NF.A.1

It rebounds to 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.

#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

Review

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

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

Standards · min grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Finding 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.

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