A cubical box dimension `L = 5//4` m starts moving with an acceleration `veca = 0.5 ms^(-2) hati` form the state of rest. At the same time, a stone is thrown form the origin with velocity `vecV = v_1hati + v_2hatj - v_3hatk` with respect to earth. Acceleration due to gravity `vecg = 10ms^(-2)(-hatj)`. The stone just touches the roof of box and finally falls at the diagonally opposite point. then:

Options

A. `v_1 = 3/2`
B. `v_2 = 5`
C. `v_3 = 5/4`
D. `v_3 = 5/2`

Explanation

To solve the problem step by step, we will analyze the motion of both the cubical box and the stone.

### Step 1: Determine the maximum height reached by the stone
The stone is thrown from the origin with an initial vertical velocity \( u_y \) and it touches the roof of the box. The maximum height \( H \) that the stone reaches is equal to the side length of the box, which is \( L = \frac{5}{4} \) m.

Using the equation of motion for vertical displacement:
\[ H = \frac{u_y^2}{2g} \]
where \( g = 10 \, \text{m/s}^2 \).

Substituting \( H \):
\[ \frac{5}{4} = \frac{u_y^2}{2 \times 10} \]
\[ \frac{5}{4} = \frac{u_y^2}{20} \]
Multiplying both sides by 20:
\[ u_y^2 = 20 \times \frac{5}{4} = 25 \]
Taking the square root:
\[ u_y = 5 \, \text{m/s} \]

### Step 2: Calculate the time of flight
The time of flight \( t \) until the stone reaches its maximum height and falls back down can be calculated using:
\[ t = \frac{2u_y}{g} \]
Substituting the values:
\[ t = \frac{2 \times 5}{10} = 1 \, \text{s} \]

### Step 3: Determine the horizontal motion of the stone
During the time of flight, the stone will also move horizontally. The horizontal displacement is given by:
\[ \text{Displacement in x-direction} = u_x \cdot t + \frac{1}{2} a_x t^2 \]
The box is accelerating with \( a_x = 0.5 \, \text{m/s}^2 \). Since the box starts from rest, we can denote the initial horizontal velocity of the stone as \( u_x \).

Substituting the known values:
\[ \text{Displacement in x-direction} = u_x \cdot 1 + \frac{1}{2} \cdot 0.5 \cdot (1)^2 \]
\[ = u_x + \frac{1}{4} \]

### Step 4: Determine the displacement of the box
The displacement of the box in the x-direction after 1 second is:
\[ \text{Displacement of box} = \frac{1}{2} a_x t^2 = \frac{1}{2} \cdot 0.5 \cdot (1)^2 = \frac{1}{4} \, \text{m} \]

### Step 5: Set the displacements equal
Since the stone just touches the roof of the box and falls at the diagonally opposite point, the horizontal displacement of the stone must equal the displacement of the box:
\[ u_x + \frac{1}{4} = \frac{5}{4} \]
Solving for \( u_x \):
\[ u_x = \frac{5}{4} - \frac{1}{4} = 1 \, \text{m/s} \]

### Step 6: Determine the z-component of the velocity
Since there is no acceleration in the z-direction, the z-component of the initial velocity \( u_z \) remains constant. The stone is thrown from the origin, so we can assume:
\[ u_z = v_3 \]
The stone falls down under gravity, so its velocity in the z-direction when it touches the roof will be negative.

### Final Result
We have:
- \( u_x = 1 \, \text{m/s} \)
- \( u_y = 5 \, \text{m/s} \)
- \( u_z = v_3 \)