Question

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 \( r \times h \) meters and it bounces indefinitely. Consider the value of \( h = 5 \) meters and \( r = \frac{1}{3} \). The total vertical distance (up and down) travelled (in meters) by the ball is ...........

Hint:

The total distance for a bouncing ball can be calculated using the sum of a geometric series. Consider both upward and downward distances for each bounce.

Answer 1

Correct Answer: 10

Step 1: Understanding the motion. 
The ball is dropped from a height \( h = 5 \) meters, and each time it bounces, it travels \( r \times h \) meters. The total vertical distance travelled by the ball includes both the downward and upward distances for each bounce. The total distance is the sum of the following series: 

- The ball initially falls \( h = 5 \) meters. 

- On the first bounce, it rises \( r \times h = \frac{1}{3} \times 5 = \frac{5}{3} \) meters and then falls the same distance. 

- On the second bounce, it rises \( r^2 \times h = \left(\frac{1}{3}\right)^2 \times 5 = \frac{5}{9} \) meters and then falls the same distance, and so on.

Step 2: Calculating the total distance. 
The total vertical distance is the sum of the initial fall, the upward and downward distances for each subsequent bounce: \[ \text{Total distance} = h + 2 \left( \frac{5}{3} + \frac{5}{9} + \frac{5}{27} + \dots \right). \]

Step 3: Summing the infinite series. 
The series is a geometric series with the first term \( \frac{5}{3} \) and common ratio \( \frac{1}{3} \). The sum of the infinite geometric series is given by: \[ S = \frac{a}{1 - r} = \frac{\frac{5}{3}}{1 - \frac{1}{3}} = \frac{\frac{5}{3}}{\frac{2}{3}} = \frac{5}{2}. \]

Step 4: Conclusion. 
The total vertical distance travelled is: \[ \text{Total distance} = 5 + 2 \times \frac{5}{2} = 5 + 5 = 10 \, \text{meters}. \] The total vertical distance travelled by the ball is \( \boxed{10} \, \text{meters} \).