A bouncing ball is dropped from an initial height of h meters above a flat surface. Each time the ball hits the surface, it rebounds a distance r × h meters and it bounces indefinitely. Consider the value of h = 5 meters and r = 1/3. The total vertical distance (up and down) travelled (in meters) by the ball is ______.
Bouncing Ball Total Vertical Distance: Infinite Series Solution (h=5m, r=1/3)
The total vertical distance traveled by the ball is 10 meters. This classic bouncing ball problem models an infinite geometric series, where the initial drop is 5 meters, followed by repeated up-and-down bounces each covering 2 × r × h distance with common ratio r = 1/3. The sum converges because |r| < 1.
Problem Breakdown
A ball drops from height h = 5 m, hits the surface, and rebounds to r × h = 5/3 m, then falls 5/3 m, rebounds to 5/9 m, and so on indefinitely[web:1][web:5]. Total distance includes one initial downward trip plus infinite pairs of up and down after each bounce. The series is D = h + 2hr + 2hr² + 2hr³ + ⋯.
Series Solution
The initial drop contributes h = 5 m. The bouncing series is 2h ∑(n=1→∞) rⁿ, a geometric series with first term a = 2hr and ratio r. Sum ∑(n=1→∞) rⁿ = r/(1-r) for |r| < 1, so total bouncing distance is 2h · r/(1-r)[web:1][web:15]. Thus, D = h + 2h · r/(1-r) = h(1 + 2r/(1-r)).
Numerical Calculation
Substitute h = 5, r = 1/3:
r/(1-r) = (1/3)/(2/3) = 1/2,
bouncing distance 2 × 5 × 1/2 = 5 m.
Total D = 5 + 5 = 10 m
Partial sums confirm: after 1st bounce 5 + 2 × 5/3 ≈ 8.33 m; after 2nd ≈ 9.26 m; approaches 10 m.
Introduction to Bouncing Ball Total Vertical Distance
The bouncing ball total vertical distance problem is a staple in physics and mathematics, modeling real-world motion with infinite geometric series. Dropped from h=5 meters with rebound ratio r=1/3, the ball travels indefinitely but converges to a finite distance—ideal for CSIR NET Life Sciences aspirants tackling physics applications in biology contexts like biomechanics. This SEO-optimized guide breaks down the calculation, series formula, and exam tips.
Step-by-Step Physics Analysis
- Initial Drop: Ball falls 5 m straight down[web:5].
- First Bounce: Rebounds
5/3m up, then falls5/3m down (total2 × 5/3 ≈ 3.33m). - Subsequent Bounces: Heights form GP:
5/3, 5/9, 5/27, …; each contributes 2 × height.
No options provided in query, but common MCQ traps include forgetting the factor of 2 (up+down), summing only downs, or using finite bounces—yielding ~7.5 m or 8.33 m instead of 10 m.
Mathematical Derivation
Total distance D = h + ∑(n=1→∞) 2 h rⁿ = h + 2hr ∑(n=0→∞) rⁿ = h + 2hr · 1/(1-r).
For r=1/3: D = 5(1 + 2·(1/3)/(1-1/3)) = 5(1 + (2/3)/(2/3)) = 5(1+1) = 10 m.
Exam Relevance for CSIR NET
CSIR NET problems often test convergence (|r|<1) and series sums in biotechnology contexts like particle tracking[user-information]. Practice variations: change r to 0.8 (D=42m) or finite bounces. Key takeaway: Always account for both up and down paths.
