Vinay dropped a ball from his terrace, whose height is 50 meters. If the…

When a ball is dropped from a height h and, after every bounce off the ground, rebounds to a fraction r of the height it fell from, the total distance it travels equals the initial drop plus twice the sum of every subsequent rebound height — because each rebound is travelled once upward and once downward. Using the sum of an infinite geometric progression, S∞ = a/(1 − r) for |r| < 1, this total distance works out to h + 2·h·r/(1 − r).

1. Here the initial height h = 50 m, and the ball reaches half of its previous height after each bounce, so r = 1/2.

2. The rebound heights form a geometric progression: 25 m, 12.5 m, 6.25 m, … with first term a = 25 and common ratio r = 1/2.

3. Sum of this infinite geometric progression = a/(1 − r) = 25/(1 − 1/2) = 25/(1/2) = 50 m.

4. Since every rebound is travelled both upward and downward, the bounces together add 2 × 50 = 100 m to the total distance.

5. Total distance = initial drop + distance from all the bounces = 50 + 100 = 150 m.

Cross-check: writing out the first few terms explicitly — 50 (initial fall) + 2(25) + 2(12.5) + 2(6.25) + … — the partial sums keep approaching 150 m as more bounces are added, confirming the closed-form result.