A stone is projected from level ground with speed u and ann at angle `theta` with horizontal. Somehow the acceleration due to gravity (g) becomes double (that is 2g) immediately after the stone reaches the maximum height and remains same thereafter. Assume direction of acceleration due to gravity always vertically downwards.
Q. The total time of flight of particle is:

Options

A. `(3)/(2)(usintheta)/(g)`
B. `(usintheta)/(g)(1+(1)/(sqrt2))`
C. `(2usintheta)/(g)`
D. `(usintheta)/(g)(2+(1)/(sqrt2))`

Explanation

To solve the problem, we need to find the total time of flight of a stone projected from the ground at an angle \( \theta \) with an initial speed \( u \), considering that the acceleration due to gravity doubles after reaching the maximum height.

### Step-by-Step Solution:

1. **Resolve the Initial Velocity:**
The initial velocity \( u \) can be resolved into two components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

2. **Time to Reach Maximum Height (\( t_1 \)):**
At maximum height, the vertical component of the velocity becomes zero. Using the first equation of motion:
\[ v_y = u_y - g t_1 \]
At maximum height, \( v_y = 0 \):
\[ 0 = u \sin \theta - g t_1 \]
Rearranging gives:
\[ t_1 = \frac{u \sin \theta}{g} \]

3. **Acceleration After Maximum Height:**
After reaching the maximum height, the acceleration due to gravity becomes \( 2g \).

4. **Time to Descend from Maximum Height to Ground (\( t_2 \)):**
For the downward motion, we can use the first equation of motion again. The initial velocity at the maximum height is \( 0 \) (since it has just reached the peak), and the acceleration is \( 2g \):
\[ v_y = 0 + (-2g) t_2 \]
The final velocity just before hitting the ground will be equal to the initial vertical velocity \( u \sin \theta \) (as it will have the same magnitude but opposite direction):
\[ -u \sin \theta = -2g t_2 \]
Rearranging gives:
\[ t_2 = \frac{u \sin \theta}{2g} \]

5. **Total Time of Flight (\( T \)):**
The total time of flight is the sum of the time to reach maximum height and the time to descend:
\[ T = t_1 + t_2 \]
Substituting the values we found:
\[ T = \frac{u \sin \theta}{g} + \frac{u \sin \theta}{2g} \]
To combine these fractions, we can take a common denominator:
\[ T = \frac{u \sin \theta}{g} \left(1 + \frac{1}{2}\right) = \frac{u \sin \theta}{g} \cdot \frac{3}{2} \]
Therefore, the total time of flight is:
\[ T = \frac{3u \sin \theta}{2g} \]

### Final Answer:
The total time of flight of the particle is:
\[ T = \frac{3u \sin \theta}{2g} \]