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 parts: the ascent to the maximum height and the descent from the maximum height back to the ground.

### Step 1: Analyze the upward motion to maximum height
When the ball is projected with an initial velocity \( u \) at an angle \( \theta \), the vertical component of the initial velocity \( u_y \) is given by:
\[ u_y = u \sin \theta \]

At the same time, there is an upward electric field \( E \) acting on the ball, which creates a downward force due to the charge \( q \). The net downward acceleration \( a_y \) during the upward motion is:
\[ a_y = g - \frac{qE}{m} \]
where \( g \) is the acceleration due to gravity.

The time \( t_1 \) taken to reach the maximum height can be calculated using the formula:
\[ t_1 = \frac{u_y}{a_y} = \frac{u \sin \theta}{g - \frac{qE}{m}} \]

### Step 2: Analyze the downward motion from maximum height to ground
Once the ball reaches the maximum height, the electric field is switched off, and the ball falls back to the ground under the influence of gravity alone. The time \( t_2 \) taken to fall from the maximum height back to the ground can be calculated using the formula for free fall:
\[ t_2 = \sqrt{\frac{2h}{g}} \]
where \( h \) is the maximum height reached. The maximum height \( h \) can be found using:
\[ h = \frac{u_y^2}{2g} = \frac{(u \sin \theta)^2}{2g} \]
Thus, substituting \( h \) into the equation for \( t_2 \):
\[ t_2 = \sqrt{\frac{2 \cdot \frac{(u \sin \theta)^2}{2g}}{g}} = \frac{u \sin \theta}{g} \]

### Step 3: Compare \( t_1 \) and \( t_2 \)
From the above calculations, we have:
\[ t_1 = \frac{u \sin \theta}{g - \frac{qE}{m}} \quad \text{and} \quad t_2 = \frac{u \sin \theta}{g} \]

Since \( g > \frac{qE}{m} \) (given \( mg > qE \)), it follows that:
\[ g - \frac{qE}{m} < g \]
This implies:
\[ t_1 < t_2 \]

### Conclusion
Thus, the time of flight to the maximum height \( t_1 \) is less than the time of flight from maximum height to the ground \( t_2 \):
\[ t_1 < t_2 \]