A particle of mass 2 kg charge 1 mC is projected vertially with velocity k`10 ms^-1`. There is as uniform horizontal electric field of `10^4N//C,` then

Options

A. the horizontal range of the particle is `10 m`
B. the time of flight of the particle is `2s`
C. the maximum heighty reached is `5m`
D. the horizontal range of the particle is `5m`

Explanation

To solve the problem step by step, we will analyze the motion of the particle under the influence of both vertical and horizontal components.

### Step 1: Identify the given parameters
- Mass of the particle (m) = 2 kg
- Charge of the particle (q) = 1 mC = \(1 \times 10^{-3}\) C
- Initial vertical velocity (u_y) = 10 m/s
- Uniform horizontal electric field (E) = \(10^4\) N/C
- Acceleration due to gravity (g) = 10 m/s² (assuming standard value for simplicity)

### Step 2: Calculate the maximum height reached by the particle
The maximum height (H) can be calculated using the formula:
\[ H = \frac{u_y^2}{2g} \]
Substituting the values:
\[ H = \frac{(10)^2}{2 \times 10} = \frac{100}{20} = 5 \text{ m} \]

### Step 3: Calculate the time of flight (time period) for the vertical motion
The total time of flight (T) can be calculated using the formula:
\[ T = \frac{2u_y}{g} \]
Substituting the values:
\[ T = \frac{2 \times 10}{10} = 2 \text{ s} \]

### Step 4: Calculate the electric force acting on the particle
The electric force (F) can be calculated using:
\[ F = qE \]
Substituting the values:
\[ F = (1 \times 10^{-3} \text{ C}) \times (10^4 \text{ N/C}) = 10 \text{ N} \]

### Step 5: Calculate the acceleration due to the electric force
The acceleration (a) in the horizontal direction can be calculated using Newton's second law:
\[ a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{10 \text{ N}}{2 \text{ kg}} = 5 \text{ m/s}^2 \]

### Step 6: Calculate the horizontal displacement (range)
The horizontal displacement (x) can be calculated using the formula:
\[ x = \frac{1}{2} a T^2 \]
Substituting the values:
\[ x = \frac{1}{2} \times 5 \times (2)^2 = \frac{1}{2} \times 5 \times 4 = 10 \text{ m} \]

### Summary of Results
- Maximum height (H) = 5 m
- Time of flight (T) = 2 s
- Horizontal range (x) = 10 m

### Final Answer
- Maximum height: 5 m
- Time period: 2 s
- Horizontal range: 10 m