A charged ball of mass m and charge q is projected from the ground with an initail velocity u and angle of projection `theta` at the time of projection vertically upward electric field E is switched on. When it reaches the maximum height electric filed is switched off and a horizontal electric feild of same magnitude E is switched on. Then answer the following question (Assume mg> qE)
Q If the time of flight to the maximum height is `t_1` and the time of flight maximum height to the ground is `t_2` then

Options

A. `t_1=t_2`
B. `t_1ltt_2`
C. `t_1gtt_2`
D. none of these

Explanation

To solve the problem, we need to analyze the motion of the charged ball in two phases: the upward motion under the influence of gravity and the electric field, and the downward motion after the electric field is switched.

### Step-by-Step Solution:

1. **Determine the time to reach maximum height (t₁)**:
The time taken to reach the maximum height can be calculated using the formula:
\[ t_1 = \frac{u \sin \theta}{g + \frac{qE}{m}} \]
Here, \( u \) is the initial velocity, \( \theta \) is the angle of projection, \( g \) is the acceleration due to gravity, \( q \) is the charge of the ball, \( E \) is the electric field strength, and \( m \) is the mass of the ball.

2. **Determine the time of flight from maximum height to the ground (t₂)**:
After reaching the maximum height, the electric field is switched off, and the ball falls under the influence of gravity alone. The time taken to fall from maximum height back to the ground can be calculated using the formula:
\[ t_2 = \sqrt{\frac{2h}{g}} \]
where \( h \) is the maximum height reached by the ball. The maximum height can be calculated using:
\[ h = \frac{(u \sin \theta)^2}{2(g + \frac{qE}{m})} \]
Thus, substituting \( h \) into the equation for \( t_2 \):
\[ t_2 = \sqrt{\frac{2 \cdot \frac{(u \sin \theta)^2}{2(g + \frac{qE}{m})}}{g}} = \frac{u \sin \theta}{\sqrt{g(g + \frac{qE}{m})}} \]

3. **Relate t₁ and t₂**:
Now, we can express the relationship between \( t_1 \) and \( t_2 \):
\[ t_1 = \frac{u \sin \theta}{g + \frac{qE}{m}}, \quad t_2 = \frac{u \sin \theta}{\sqrt{g(g + \frac{qE}{m})}} \]
To find a relationship between \( t_1 \) and \( t_2 \), we can take the ratio:
\[ \frac{t_1}{t_2} = \frac{\sqrt{g(g + \frac{qE}{m})}}{g + \frac{qE}{m}} \]

### Final Answer:
The time of flight to the maximum height \( t_1 \) and the time of flight from maximum height to the ground \( t_2 \) can be expressed in terms of the initial velocity, angle of projection, electric field, and gravitational acceleration.