A ball is projected from the ground with speed u at an angle `alpha` with horizontal. It collides with a wall at a distance a from the point of projection and returns to its original position. Find the coefficient of restitution between the ball and the wall.

Explanation

To find the coefficient of restitution (E) between the ball and the wall, we can follow these steps:

### Step 1: Understand the Motion of the Ball
The ball is projected with an initial speed \( u \) at an angle \( \alpha \) with the horizontal. It travels a parabolic path, collides with the wall at a horizontal distance \( a \), and then returns to its original position.

### Step 2: Resolve the Initial Velocity
The initial velocity \( u \) can be resolved into horizontal and vertical components:
- Horizontal component: \( u_x = u \cos \alpha \)
- Vertical component: \( u_y = u \sin \alpha \)

### Step 3: Time to Reach the Wall
The time \( t_i \) taken to reach the wall can be calculated using the horizontal motion:
\[ t_i = \frac{a}{u \cos \alpha} \]

### Step 4: Total Time of Flight
The total time of flight \( T \) for a projectile that returns to the same vertical level is given by:
\[ T = \frac{2u \sin \alpha}{g} \]
where \( g \) is the acceleration due to gravity.

### Step 5: Time After Collision
Let \( t_b \) be the time taken to return to the original position after the collision. The total time of flight can be expressed as:
\[ T = t_i + t_b \]
Thus,
\[ t_b = T - t_i = \frac{2u \sin \alpha}{g} - \frac{a}{u \cos \alpha} \]

### Step 6: Velocity After Collision
After the collision, the ball's horizontal velocity changes due to the coefficient of restitution. The horizontal component of the velocity after the collision is:
\[ v_x = E u \cos \alpha \]
where \( E \) is the coefficient of restitution.

### Step 7: Distance Traveled After Collision
The distance traveled after the collision is again \( a \), and can be expressed in terms of the new velocity:
\[ t_b = \frac{a}{E u \cos \alpha} \]

### Step 8: Equate the Two Expressions for Time
Now, equate the expressions for \( t_b \):
\[ \frac{2u \sin \alpha}{g} - \frac{a}{u \cos \alpha} = \frac{a}{E u \cos \alpha} \]

### Step 9: Solve for E
Rearranging the equation gives:
\[ \frac{2u \sin \alpha}{g} = \frac{a}{u \cos \alpha} + \frac{a}{E u \cos \alpha} \]
Factoring out \( \frac{a}{u \cos \alpha} \):
\[ \frac{2u \sin \alpha}{g} = \frac{a}{u \cos \alpha} \left(1 + \frac{1}{E}\right) \]
Cross-multiplying leads to:
\[ 2u^2 \sin \alpha \cos \alpha = \frac{a}{g} \left(1 + \frac{1}{E}\right) \]
Thus,
\[ 1 + \frac{1}{E} = \frac{2u^2 \sin \alpha \cos \alpha g}{a} \]
This simplifies to:
\[ \frac{1}{E} = \frac{2u^2 \sin \alpha \cos \alpha g}{a} - 1 \]
Finally, solving for \( E \):
\[ E = \frac{a g}{a g - 2u^2 \sin \alpha \cos \alpha} \]

### Final Result
The coefficient of restitution \( E \) is given by:
\[ E = \frac{a g}{a g - u^2 \sin 2\alpha} \]