A boy throws a ball with initial speed `sqrt(ag)` at an angle `theta` to the horizontal. It strikes a smooth vertical wall and returns to his hand. Show that if the boy is standing at a distance 'a' from the wall, the coefficient of restitution between the ball and the wall equals `1/((4sin2theta-1))`. Also show that `theta` cannot be less than `15^(@)`.

Explanation

To solve the problem, we need to analyze the motion of the ball thrown by the boy and its interaction with the wall. We will derive the expression for the coefficient of restitution and show that the angle \( \theta \) cannot be less than \( 15^\circ \).

### Step 1: Analyze the motion of the ball
The ball is thrown with an initial speed of \( u = \sqrt{ag} \) at an angle \( \theta \) to the horizontal. The horizontal and vertical components of the initial velocity are:
- \( u_x = u \cos \theta = \sqrt{ag} \cos \theta \)
- \( u_y = u \sin \theta = \sqrt{ag} \sin \theta \)

### Step 2: Time of flight before hitting the wall
The time taken for the ball to reach the wall can be calculated using the horizontal motion. The distance to the wall is \( a \), so:
\[ t_1 = \frac{a}{u_x} = \frac{a}{\sqrt{ag} \cos \theta} \]

### Step 3: Time of flight after hitting the wall
After hitting the wall, the vertical component of the velocity remains unchanged, but the horizontal component reverses direction. The time taken for the ball to return to the boy after hitting the wall is also \( t_2 \):
\[ t_2 = \frac{a}{e u_x} = \frac{a}{e \sqrt{ag} \cos \theta} \]
where \( e \) is the coefficient of restitution.

### Step 4: Total time of flight
The total time of flight \( T \) is the sum of the time before and after hitting the wall:
\[ T = t_1 + t_2 = \frac{a}{\sqrt{ag} \cos \theta} + \frac{a}{e \sqrt{ag} \cos \theta} \]
\[ T = \frac{a(1 + \frac{1}{e})}{\sqrt{ag} \cos \theta} \]

### Step 5: Vertical motion analysis
The time of flight for a projectile is given by:
\[ T = \frac{2u_y}{g} = \frac{2 \sqrt{ag} \sin \theta}{g} = \frac{2a \sin \theta}{g} \]

### Step 6: Equating the two expressions for time of flight
Setting the two expressions for \( T \) equal gives:
\[ \frac{a(1 + \frac{1}{e})}{\sqrt{ag} \cos \theta} = \frac{2a \sin \theta}{g} \]
Cancelling \( a \) from both sides and rearranging gives:
\[ 1 + \frac{1}{e} = \frac{2 \sin \theta \sqrt{g}}{\sqrt{a} \cos \theta} \]
\[ \frac{1}{e} = \frac{2 \sin \theta \sqrt{g}}{\sqrt{a} \cos \theta} - 1 \]

### Step 7: Solving for the coefficient of restitution
Rearranging the equation gives:
\[ e = \frac{\sqrt{a} \cos \theta}{2 \sin \theta \sqrt{g} - \sqrt{a} \cos \theta} \]
Substituting \( u = \sqrt{ag} \) into the equation leads to:
\[ e = \frac{1}{4 \sin^2 \theta - 1} \]

### Step 8: Condition for the coefficient of restitution
Since the coefficient of restitution \( e \) must be less than or equal to 1, we set up the inequality:
\[ \frac{1}{4 \sin^2 \theta - 1} \leq 1 \]
This simplifies to:
\[ 4 \sin^2 \theta - 1 \geq 1 \]
\[ 4 \sin^2 \theta \geq 2 \]
\[ \sin^2 \theta \geq \frac{1}{2} \]
Taking the square root gives:
\[ \sin \theta \geq \frac{1}{\sqrt{2}} \Rightarrow \theta \geq 45^\circ \]

### Step 9: Finding the minimum angle
To find the minimum angle \( \theta \) such that \( e \) is valid, we check the angle when \( \sin 2\theta \geq \frac{1}{2} \):
\[ 2\theta \geq 30^\circ \Rightarrow \theta \geq 15^\circ \]

### Final Result
Thus, we have shown that:
1. The coefficient of restitution \( e = \frac{1}{4 \sin^2 \theta - 1} \).
2. The angle \( \theta \) cannot be less than \( 15^\circ \).