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Exercise 5.2 · Q17

Q.After striking a floor a certain ball rebounds 45\dfrac{4}{5}th of the height from which it has fallen. If the ball is dropped from a height of 240 cm, find the total distance the ball travels before coming to rest.

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The ball falls once, then each rebound-and-fall adds a further 2×2\times(previous drop height); sum the resulting infinite G.P.

Total distance =h+2(hr+hr2+hr3+⋯ )=h+2hr1−r=h+2\left(hr+hr^2+hr^3+\cdots\right)=h+\dfrac{2hr}{1-r}, where hh = initial drop height, rr = rebound fraction.

  1. h=240h=240 cm, r=45r=\dfrac45 (fraction of height regained on each bounce).
  2. After the first fall (240240 cm down), the ball rebounds to 240r240r and falls the same distance again — so each subsequent bounce contributes 2×2\times that height (up + down), forming an infinite G.P. of ratio rr.
  3. Total distance =h+2hr+2hr2+2hr3+⋯=h+2hr1−r=h+2hr+2hr^2+2hr^3+\cdots=h+\dfrac{2hr}{1-r}. …

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