Operations & Word Problems

Problem

Bouncing ball rises a fraction of its fall

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.
FractionsOperations
Your answer
m
How to solve
Strategy Draw a Diagram — 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.
1STEP 1

First fall

The ball drops the full starting height of 150 m.

150 m
2STEP 2

First rise (first bounce up)

It rebounds to 2/5 of the 150 m fall — divide by 5, then multiply by 2, to get 60 m.

150 ÷ 5 = 30, 30 × 2 = 60
3STEP 3

Second fall

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

60 m
4STEP 4

Second rise (second bounce up)

It rebounds to 2/5 of the 60 m it just fell — divide by 5, then multiply by 2, to get 24 m.

60 ÷ 5 = 12, 12 × 2 = 24
5STEP 5

Add all four legs

Add the first fall, first rise, second fall, and second rise: 150 + 60 + 60 + 24 = 294 m.

150 + 60 + 60 + 24 = 294
Answer
294 m
150 + 60 + 60 + 24 = 294
The drop then each rebound shrinks in turn (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.
Takeaway

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

  • First fall
  • First rise (first bounce up)
  • Second fall
  • Second rise (second bounce up)
  • Add all four legs
Where next?
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