Analysis:
To determine the angle theta for which the horizontal range inside the box is maximum, we need to consider the motion of the particle with respect to both the box and the ground frame. Let's break down the problem into smaller parts and analyze them step by step.

1. Motion of the box:
The box is moving on a horizontal floor with constant acceleration a = g (acceleration due to gravity). Since the acceleration is constant, the box will undergo uniformly accelerated motion.

2. Motion of the particle relative to the box:
The particle is projected inside the box with velocity u and angle theta with the horizontal from the box frame. We can resolve the initial velocity into horizontal and vertical components as follows:
- Horizontal component: u * cos(theta)
- Vertical component: u * sin(theta)

3. Motion of the particle relative to the ground:
To analyze the motion of the particle relative to the ground, we need to consider the velocity of the box as well. Since the box is moving with constant acceleration, the velocity of the box will be v = u + at, where t is the time.

4. Horizontal range:
The horizontal range is the distance traveled by the particle in the horizontal direction before hitting the ground. We can calculate the time of flight using the vertical motion of the particle relative to the ground. The time of flight (T) can be calculated using the equation:
- T = 2 * (u * sin(theta)) / g

The horizontal range (R) can be calculated using the horizontal component of the velocity and the time of flight:
- R = (u * cos(theta)) * T

5. Maximizing the horizontal range:
To maximize the horizontal range, we need to find the angle theta that gives the maximum value of R. We can achieve this by differentiating R with respect to theta and equating it to zero. Solving this equation will give us the value of theta for which the horizontal range is maximum.

6. Solution:
Differentiating R with respect to theta:
- dR/d(theta) = (u * cos(theta)) * dT/d(theta) + (-u * sin(theta)) * T

Setting dR/d(theta) = 0:
- (u * cos(theta)) * dT/d(theta) + (-u * sin(theta)) * T = 0

Substituting the value of T from the equation of time of flight:
- (u * cos(theta)) * (2 * u * cos(theta) / g) + (-u * sin(theta)) * (2 * (u * sin(theta)) / g) = 0

Simplifying the equation:
- 2 * (u^2 * cos^2(theta) / g) - 2 * (u^2 * sin^2(theta) / g) = 0

Dividing both sides by 2 * (u^2 / g):
- cos^2(theta) - sin^2(theta) = 0

Using the identity cos^2(theta) - sin^2(theta) = cos(2 * theta):
- cos(2 * theta) = 0

Solving for theta:
- 2 * theta = pi/2
- theta = pi/4

Therefore, the angle theta for which the horizontal range inside the