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Worked Examples · Example 38

Q.A ball is dropped from a height of 90 feet and always rebounds one-third the distance from which it falls. Find the total vertical distance the ball travelled when it hits the ground for the 3rd time. Also find the total distance travelled by the ball before it comes to rest.

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The ball's drop and each rebound-and-fall form a pattern of decreasing heights (a GP with ratio 13\tfrac13); we track the running total up to the third ground-hit, then sum the whole infinite series for the total distance before rest.

For a ball dropped from height hh that rebounds a fraction rr of the previous fall each time, each bounce after the first contributes 2×(rebound height)2\times(\text{rebound height}) (up then down). Total distance =h+2rh+2r2h+…=h+2rh1−r= h + 2r h + 2r^2h+\ldots = h + \dfrac{2rh}{1-r} using the GP sum S∞=a1−rS_\infty=\dfrac{a}{1-r}.

  1. Initial drop: the ball falls 9090 ft and hits the ground for the 1st time. Distance so far =90=90 ft.
  2. First rebound: it bounces up to 13×90=30\tfrac13 \times 90 = 30 ft, then falls back down 3030 ft, hitting the ground for the 2nd time. This bounce adds 30+30=6030+30=60 ft. Running total =90+60=150=90+60=150 ft.
  3. Second rebound: it bounces up to 13×30=10\tfrac13\times 30=10 ft, then falls back down 1010 ft, hitting the ground for the 3rd time. This bounce adds 10+10=2010+10=20 ft. Running total =150+20=170=150+20=170 ft.
  4. Distance travelled when it hits the ground the 3rd time =90+2(30)+2(10)=90+60+20=170=90+2(30)+2(10)=90+60+20=170 ft. …

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