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^2sin2theta)/(g)`
B. `(u^2sin2theta)/(2g)(1+(1)/(sqrt2))`
C. `(u^2)/(g)sin2theta`
D. `(u^2sin2theta)/(2g)(2+(1)/(sqrt2))`

Explanation

To solve the problem, we will break it down into several steps:

### Step 1: Determine the time to reach maximum height
The stone is projected with an initial speed \( u \) at an angle \( \theta \) with the horizontal. The vertical component of the initial velocity is given by:
\[ u_y = u \sin \theta \]
The time taken to reach the maximum height \( t_1 \) can be calculated using the equation:
\[ v_y = u_y - g t_1 \]
At maximum height, the vertical velocity \( v_y \) becomes 0. Thus, we have:
\[ 0 = u \sin \theta - g t_1 \]
From this, we can solve for \( t_1 \):
\[ t_1 = \frac{u \sin \theta}{g} \]

### Step 2: Calculate the horizontal distance covered until maximum height
The horizontal component of the initial velocity is:
\[ u_x = u \cos \theta \]
The horizontal distance \( r_1 \) covered during the time \( t_1 \) is:
\[ r_1 = u_x t_1 = (u \cos \theta) \left(\frac{u \sin \theta}{g}\right) = \frac{u^2 \sin \theta \cos \theta}{g} \]
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we can rewrite \( r_1 \):
\[ r_1 = \frac{u^2 \sin 2\theta}{2g} \]

### Step 3: Calculate the maximum height reached
Using the kinematic equation for vertical motion, the maximum height \( h \) can be calculated as:
\[ h = u_y t_1 - \frac{1}{2} g t_1^2 \]
Substituting \( u_y \) and \( t_1 \):
\[ h = (u \sin \theta) \left(\frac{u \sin \theta}{g}\right) - \frac{1}{2} g \left(\frac{u \sin \theta}{g}\right)^2 \]
Simplifying this gives:
\[ h = \frac{u^2 \sin^2 \theta}{g} - \frac{1}{2} g \cdot \frac{u^2 \sin^2 \theta}{g^2} = \frac{u^2 \sin^2 \theta}{g} - \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin^2 \theta}{2g} \]

### Step 4: Calculate the time taken to fall from maximum height to ground
After reaching maximum height, the acceleration due to gravity becomes \( 2g \). The time \( t_2 \) taken to fall from height \( h \) to the ground can be calculated using:
\[ h = \frac{1}{2} (2g) t_2^2 \]
Substituting \( h \):
\[ \frac{u^2 \sin^2 \theta}{2g} = g t_2^2 \]
Solving for \( t_2^2 \):
\[ t_2^2 = \frac{u^2 \sin^2 \theta}{4g^2} \]
Taking the square root:
\[ t_2 = \frac{u \sin \theta}{2g} \]

### Step 5: Calculate the horizontal distance covered during the fall
The horizontal distance \( r_2 \) covered during the time \( t_2 \) is:
\[ r_2 = u_x t_2 = (u \cos \theta) \left(\frac{u \sin \theta}{2g}\right) = \frac{u^2 \sin \theta \cos \theta}{2g} \]
Again, using the identity:
\[ r_2 = \frac{u^2 \sin 2\theta}{4g} \]

### Step 6: Calculate the total horizontal range
The total horizontal range \( R \) is the sum of \( r_1 \) and \( r_2 \):
\[ R = r_1 + r_2 = \frac{u^2 \sin 2\theta}{2g} + \frac{u^2 \sin 2\theta}{4g} \]
Finding a common denominator:
\[ R = \frac{2u^2 \sin 2\theta}{4g} + \frac{u^2 \sin 2\theta}{4g} = \frac{3u^2 \sin 2\theta}{4g} \]

### Final Result
Thus, the horizontal range of the stone is:
\[ R = \frac{3u^2 \sin 2\theta}{4g} \]