A projectile is fired with a speed u at an angle `theta` above a horizontal field. The coefficient of restitution of collision between the projectile and the field is e. How far from the starting point does the projectile makes its second collision with the field?

Explanation

To solve the problem of finding how far from the starting point a projectile makes its second collision with the field after being fired with speed \( u \) at an angle \( \theta \) and having a coefficient of restitution \( e \), we can follow these steps:

### Step 1: Determine the initial velocities in the x and y directions
The initial velocity components can be calculated as:
- \( u_x = u \cos \theta \) (horizontal component)
- \( u_y = u \sin \theta \) (vertical component)

### Step 2: Calculate the time of flight until the first collision
The time of flight until the projectile hits the ground for the first time can be calculated using the formula for vertical motion:
\[ t_1 = \frac{2u_y}{g} = \frac{2u \sin \theta}{g} \]

### Step 3: Calculate the range after the first collision
The range \( R_1 \) for the first collision can be calculated using the horizontal motion:
\[ R_1 = u_x \cdot t_1 = (u \cos \theta) \cdot \left(\frac{2u \sin \theta}{g}\right) = \frac{2u^2 \sin \theta \cos \theta}{g} \]
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we can simplify this to:
\[ R_1 = \frac{u^2 \sin 2\theta}{g} \]

### Step 4: Determine the velocity after the first collision
After the first collision, the vertical component of the velocity changes due to the coefficient of restitution \( e \):
- The final vertical velocity \( v_y \) after the first collision is given by:
\[ v_y = -e u_y = -e (u \sin \theta) \]
The horizontal component remains unchanged:
\[ v_x = u_x = u \cos \theta \]

### Step 5: Calculate the time of flight until the second collision
The time of flight until the projectile hits the ground for the second time can be calculated using the new vertical velocity:
\[ t_2 = \frac{2|v_y|}{g} = \frac{2e u \sin \theta}{g} \]

### Step 6: Calculate the range after the second collision
The range \( R_2 \) for the second collision can be calculated similarly:
\[ R_2 = v_x \cdot t_2 = (u \cos \theta) \cdot \left(\frac{2e u \sin \theta}{g}\right) = \frac{2e u^2 \sin \theta \cos \theta}{g} \]
Again, using the identity \( \sin 2\theta \):
\[ R_2 = \frac{e u^2 \sin 2\theta}{g} \]

### Step 7: Calculate the total distance from the starting point
The total distance from the starting point after the second collision is the sum of the ranges:
\[ R = R_1 + R_2 = \frac{u^2 \sin 2\theta}{g} + \frac{e u^2 \sin 2\theta}{g} \]
Factoring out common terms:
\[ R = \frac{u^2 \sin 2\theta (1 + e)}{g} \]

### Final Answer
Thus, the distance from the starting point where the projectile makes its second collision with the field is:
\[ R = \frac{u^2 \sin 2\theta (1 + e)}{g} \]