A ball is dropped from a height h on a floor. The coefficient of restitution for the collision between the ball and the floor is e. The total distance covered by the ball before it comes to the rest.

Options

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

Answer

To solve the problem of finding the total distance covered by a ball dropped from a height \( h \) with a coefficient of restitution \( e \), we can follow these steps:

### Step 1: Understand the initial conditions
The ball is dropped from a height \( h \). When it hits the ground, it will bounce back to a certain height determined by the coefficient of restitution \( e \).

### Step 2: Calculate the speed just before impact
When the ball is dropped from height \( h \), it accelerates due to gravity. The speed \( u \) just before it hits the ground can be calculated using the equation of motion:
\[ u = \sqrt{2gh} \]

### Step 3: Calculate the height after the first bounce
After the first impact, the speed of the ball just after the collision is given by:
\[ v = e \cdot u \]
The maximum height \( h_1 \) reached after the first bounce can be calculated using:
\[ h_1 = \frac{v^2}{2g} = \frac{(e \cdot u)^2}{2g} = \frac{e^2 \cdot (2gh)}{2g} = e^2 h \]

### Step 4: Calculate the distance covered in the first bounce
The total distance covered during the first drop and the first bounce is:
\[ \text{Distance}_1 = h + h_1 = h + e^2 h = h(1 + e^2) \]

### Step 5: Calculate the height after subsequent bounces
After the first bounce, the ball will continue to bounce. The height after the second bounce \( h_2 \) can be calculated similarly:
\[ h_2 = e^2 h_1 = e^2 (e^2 h) = e^4 h \]
The distance covered during the second drop and bounce is:
\[ \text{Distance}_2 = h_2 + h_2 = 2h_2 = 2e^4 h \]

### Step 6: Generalize the distance for all bounces
Continuing this process, the height after the \( n \)-th bounce is:
\[ h_n = e^{2n} h \]
The distance covered during the \( n \)-th drop and bounce is:
\[ \text{Distance}_n = 2h_n = 2e^{2n} h \]

### Step 7: Sum the total distance covered
The total distance \( D \) can be expressed as:
\[ D = h(1 + e^2) + 2(e^2 h + e^4 h + e^6 h + \ldots) \]
The series \( e^2 h + e^4 h + e^6 h + \ldots \) is a geometric series with first term \( a = e^2 h \) and common ratio \( r = e^2 \). The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{e^2 h}{1 - e^2} \]
Thus, the total distance becomes:
\[ D = h(1 + e^2) + 2 \cdot \frac{e^2 h}{1 - e^2} \]
Combining these terms, we have:
\[ D = h \left( 1 + e^2 + \frac{2e^2}{1 - e^2} \right) \]

### Step 8: Final expression
Simplifying the expression:
\[ D = h \left( \frac{(1 - e^2) + e^2(1 + 2)}{1 - e^2} \right) = h \left( \frac{1 + e^2}{1 - e^2} \right) \]

### Conclusion
The total distance covered by the ball before it comes to rest is:
\[ D = h \left( \frac{1 + e^2}{1 - e^2} \right) \]