A ball falls from a height of 5m and each time it rises by 81/100 of its initial height and so on. (g=10m/`s^2`) Find average speed for long time.

Options

A. 3
B. 2.5
C. 2
D. 3.5

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Motion of the Ball
The ball is dropped from a height of \( h = 5 \, \text{m} \) and rebounds to a height that is \( k = \frac{81}{100} \) of the height from which it fell. This process continues indefinitely.

### Step 2: Calculate the Total Distance Traveled
1. The ball falls a distance of \( h = 5 \, \text{m} \).
2. It then rebounds to a height of \( h_1 = k \cdot h = \frac{81}{100} \cdot 5 = 4.05 \, \text{m} \).
3. The ball then falls back down the same distance \( h_1 \).
4. This pattern continues with each subsequent height being \( h_n = k^n \cdot h \).

The total distance \( D \) traveled by the ball can be expressed as:
\[ D = h + 2(h_1 + h_2 + h_3 + \ldots) \]
Where \( h_1, h_2, h_3, \ldots \) are the heights after each rebound.

### Step 3: Sum the Infinite Series
The heights form a geometric series:
\[ h_1 + h_2 + h_3 + \ldots = \frac{81}{100}h + \frac{81^2}{100^2}h + \frac{81^3}{100^3}h + \ldots \]
The sum of this infinite geometric series can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
Where \( a \) is the first term and \( r \) is the common ratio. Here, \( a = \frac{81}{100}h \) and \( r = \frac{81}{100} \):
\[ S = \frac{\frac{81}{100}h}{1 - \frac{81}{100}} = \frac{\frac{81}{100}h}{\frac{19}{100}} = \frac{81h}{19} \]
Thus, the total distance \( D \) becomes:
\[ D = h + 2 \cdot \frac{81h}{19} = 5 + 2 \cdot \frac{81 \cdot 5}{19} = 5 + \frac{810}{19} = \frac{95 + 810}{19} = \frac{905}{19} \, \text{m} \]

### Step 4: Calculate the Total Time Taken
1. The time taken for the first fall \( t_0 \) can be calculated using the second equation of motion:
\[ t_0 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \cdot 5}{10}} = \sqrt{1} = 1 \, \text{s} \]
2. The time taken for each rebound can be calculated similarly. For the first rebound:
\[ t_1 = 2\sqrt{\frac{2h_1}{g}} = 2\sqrt{\frac{2 \cdot 4.05}{10}} = 2\sqrt{0.81} = 2 \cdot 0.9 = 1.8 \, \text{s} \]
3. The total time \( T \) taken can be expressed as:
\[ T = t_0 + 2(t_1 + t_2 + t_3 + \ldots) \]
Where \( t_n = 2\sqrt{\frac{2h_n}{g}} \).

### Step 5: Sum the Infinite Time Series
The time for each rebound follows a similar geometric series:
\[ T = 1 + 2 \left( \sqrt{\frac{2h}{g}} \left( \frac{81}{100} + \left(\frac{81}{100}\right)^{3/2} + \ldots \right) \right) \]
Using the same formula for the sum of the series:
\[ T = 1 + 2 \cdot \sqrt{\frac{2h}{g}} \cdot \frac{\frac{81}{100}}{1 - \frac{81}{100}} = 1 + 2 \cdot \sqrt{\frac{2h}{g}} \cdot \frac{81}{19} \]

### Step 6: Calculate Average Speed
The average speed \( v_{avg} \) is given by:
\[ v_{avg} = \frac{D}{T} \]

### Final Calculation
Substituting the values of \( D \) and \( T \) into the average speed formula will yield the final result.