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 horizontal range of particle is

Options

A. `(3)/(4)(u^(2)sin2theta)/(g)`
B. `(u^(2)sin 2theta)/(2g) (1+(1)/(sqrt(2)))`
C. `(u^(2))/(g) sin 2theta`
D. `(u^(2)sin2theta)/(2g) (2+(1)/(sqrt(2)))`

Explanation

To find the horizontal range of a stone projected from level ground with speed \( u \) at an angle \( \theta \) with the horizontal, where the acceleration due to gravity becomes \( 2g \) after reaching the maximum height, we can follow these steps:

### Step 1: Calculate the time to reach maximum height
The time taken to reach the maximum height \( t_1 \) can be calculated using the formula:
\[ t_1 = \frac{u \sin \theta}{g} \]

### Step 2: Calculate the maximum height reached
The maximum height \( H \) reached by the stone can be calculated using the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

### Step 3: Calculate the time to fall from maximum height to ground
After reaching the maximum height, the stone falls under the influence of gravity \( 2g \). The time taken to fall from height \( H \) can be calculated using the second equation of motion:
\[ H = \frac{1}{2} (2g) t_2^2 \]
Solving for \( t_2 \):
\[ t_2 = \sqrt{\frac{H}{g}} = \sqrt{\frac{u^2 \sin^2 \theta}{4g^2}} = \frac{u \sin \theta}{2g} \]

### Step 4: Total time of flight
The total time of flight \( T \) is the sum of the time to reach maximum height and the time to fall back to the ground:
\[ T = t_1 + t_2 = \frac{u \sin \theta}{g} + \frac{u \sin \theta}{2g} = \frac{u \sin \theta}{g} \left(1 + \frac{1}{2}\right) = \frac{u \sin \theta}{g} \cdot \frac{3}{2} = \frac{3u \sin \theta}{2g} \]

### Step 5: Calculate the horizontal range
The horizontal range \( R \) can be calculated using the formula:
\[ R = u \cos \theta \cdot T \]
Substituting the value of \( T \):
\[ R = u \cos \theta \cdot \frac{3u \sin \theta}{2g} = \frac{3u^2 \sin \theta \cos \theta}{2g} \]
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ R = \frac{3u^2}{4g} \sin 2\theta \]

### Final Answer
The horizontal range of the particle is:
\[ R = \frac{3u^2}{4g} \sin 2\theta \]

---