A partical falls from a height `h` upon a fixed horizontal plane and rebounds. If `e` is the coefficient of restitution, the total distance travelled before rebounding has stopped is

Options

A. `h((1+e^(2))/(1-e^(2)))`
B. `h((1-e^(2))/(1+e^(2)))`
C. `h/2((1-e^(2))/(1+e^(2)))`
D. `h/2((1+e^(2))/(1-e^(2)))`

Explanation

To solve the problem of a particle falling from a height \( h \) and rebounding with a coefficient of restitution \( e \), we can follow these steps:

### Step 1: Understand the initial fall
When the particle falls from a height \( h \), it hits the ground with a certain velocity \( v \). The velocity just before impact can be calculated using the equation for free fall:
\[ v = \sqrt{2gh} \]

### Step 2: Calculate the rebound height
After the particle hits the ground, it rebounds to a height determined by the coefficient of restitution \( e \). The velocity of the particle just after the rebound is given by:
\[ u = e \cdot v \]
The height to which it rebounds can be calculated using the formula:
\[ h_1 = \frac{u^2}{2g} = \frac{(e \cdot v)^2}{2g} = \frac{e^2 v^2}{2g} \]

### Step 3: Calculate the distance traveled during the rebound
The particle will travel up to height \( h_1 \) and then fall back down the same distance. Therefore, the total distance for the first rebound is:
\[ \text{Distance for first rebound} = 2h_1 = 2 \cdot \frac{e^2 v^2}{2g} = \frac{e^2 v^2}{g} \]

### Step 4: Repeat for subsequent rebounds
For the next rebound, the velocity just after the second impact will be:
\[ u_2 = e \cdot u = e^2 \cdot v \]
The height for the second rebound will be:
\[ h_2 = \frac{(e^2 \cdot v)^2}{2g} = \frac{e^4 v^2}{2g} \]
The distance traveled during this rebound is:
\[ \text{Distance for second rebound} = 2h_2 = 2 \cdot \frac{e^4 v^2}{2g} = \frac{e^4 v^2}{g} \]

### Step 5: Generalize the pattern
Continuing this process, we can see that the distance traveled for the \( n \)-th rebound will be:
\[ \text{Distance for } n\text{-th rebound} = 2h_n = 2 \cdot \frac{(e^{2n} v^2)}{2g} = \frac{e^{2n} v^2}{g} \]

### Step 6: Calculate the total distance traveled
The total distance \( D \) traveled by the particle before it stops rebounding is the sum of the initial drop and all the rebounds:
\[ D = h + 2h_1 + 2h_2 + 2h_3 + \ldots \]
This can be expressed as:
\[ D = h + \sum_{n=1}^{\infty} \frac{e^{2n} v^2}{g} \]
Recognizing that the series is a geometric series with first term \( e^2 v^2/g \) and common ratio \( e^2 \):
\[ \sum_{n=0}^{\infty} e^{2n} = \frac{1}{1 - e^2} \quad \text{(for } |e| < 1\text{)} \]
Thus, we have:
\[ D = h + \frac{e^2 v^2}{g} \cdot \frac{1}{1 - e^2} \]

### Step 7: Substitute \( v^2 \)
Since \( v^2 = 2gh \), we substitute this into the equation:
\[ D = h + \frac{e^2 (2gh)}{g(1 - e^2)} = h + \frac{2e^2 h}{1 - e^2} \]
Factoring out \( h \):
\[ D = h \left(1 + \frac{2e^2}{1 - e^2}\right) \]

### Final Expression
Thus, the total distance traveled before rebounding has stopped is:
\[ D = h \left(\frac{1 - e^2 + 2e^2}{1 - e^2}\right) = h \left(\frac{1 + e^2}{1 - e^2}\right) \]

### Conclusion
Therefore, the total distance traveled by the particle before it stops rebounding is:
\[ D = h \cdot \frac{1 + e^2}{1 - e^2} \]