A particle is thrown vertically upward with a speed `u` from the top of a tower of height `h` from ground level, If after first impact with ground it just reaches to height `h` from ground the coeffecient of restitution for the collision is `:-`

Options

A. `1`
B. `u^(2)`
C. `sqrt((2gh)/(U^(2)+2gh))`
D. `sqrt(2gh)/(U)`

Explanation

To solve the problem, we need to find the coefficient of restitution (e) for a particle thrown vertically upward from the top of a tower. The particle reaches a height `h` after the first impact with the ground.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A particle is thrown upward with an initial speed `u` from a height `h`.
- After hitting the ground, it reaches a height `h` again.
- We need to find the coefficient of restitution (e).

2. **Finding the Velocity Just Before Impact**:
- When the particle is thrown upward, it will eventually come down and hit the ground.
- We can use the kinematic equation to find the velocity just before impact (v):
\[ v^2 = u^2 + 2gh \]
- Here, `g` is the acceleration due to gravity, and we consider downward motion as positive.

3. **Finding the Velocity Just After Impact**:
- After the impact, the particle reaches a height `h` again. At the maximum height, the final velocity (v') is 0.
- We can use the kinematic equation again to find the velocity just after impact (v'):
\[ 0 = (v')^2 - 2gh \]
Rearranging gives:
\[ (v')^2 = 2gh \implies v' = \sqrt{2gh} \]

4. **Using the Coefficient of Restitution**:
- The coefficient of restitution (e) is defined as:
\[ e = \frac{\text{velocity of separation}}{\text{velocity of approach}} = \frac{v'}{v} \]
- Substituting the values we found:
\[ e = \frac{\sqrt{2gh}}{\sqrt{u^2 + 2gh}} \]

5. **Final Expression**:
- Therefore, the coefficient of restitution is:
\[ e = \frac{\sqrt{2gh}}{\sqrt{u^2 + 2gh}} \]