STATEMENT-1: A body is projected from the ground with a velocity v at an angle with the horizontal direction, it reaches a maximum height (H) and reaches the ground after a time T = 2u sin `theta//g`
STATEMENT-2: The vertical and horizontal motions can be treated independently

Options

A. Statement-1 is True, Statement-2 is True, Statement -2 is a correct explanation for Statement-1
B. Statement-1 is True, Statement-2 is True, Statement -2 is NOT a correct explanation for Statement-1.
C. Statement - 1 is True, Statement-2 is False
D. Statement - 1 is False, Statement-2 is True.

Answer

To solve the problem, we need to analyze the motion of a projectile launched from the ground at an angle \(\theta\) with an initial velocity \(v\). We will derive the time of flight and discuss the independence of vertical and horizontal motions.

### Step-by-Step Solution:

1. **Understanding the Components of Motion:**
- When a body is projected at an angle \(\theta\), its initial velocity \(v\) can be resolved into two components:
- Horizontal component: \(v_x = v \cos \theta\)
- Vertical component: \(v_y = v \sin \theta\)

2. **Time of Flight:**
- The time of flight \(T\) for a projectile is the total time it takes for the projectile to return to the ground after being launched.
- The vertical motion can be analyzed using the equation of motion:
\[ S_y = v_y t - \frac{1}{2} g t^2 \]
where \(S_y\) is the vertical displacement (which is 0 when it returns to the ground), \(v_y\) is the initial vertical velocity, \(g\) is the acceleration due to gravity, and \(t\) is the time of flight.
- Setting \(S_y = 0\) gives:
\[ 0 = v \sin \theta \cdot T - \frac{1}{2} g T^2 \]
- Rearranging this equation leads to:
\[ g T^2 = 2 v \sin \theta \cdot T \]
- Dividing both sides by \(T\) (assuming \(T \neq 0\)):
\[ g T = 2 v \sin \theta \]
- Therefore, the time of flight \(T\) is:
\[ T = \frac{2 v \sin \theta}{g} \]

3. **Maximum Height:**
- The maximum height \(H\) can be calculated using the vertical motion equations. At maximum height, the vertical velocity becomes zero.
- Using the equation:
\[ v_y^2 = u_y^2 - 2gH \]
where \(v_y = 0\) at maximum height, we have:
\[ 0 = (v \sin \theta)^2 - 2gH \]
- Rearranging gives:
\[ H = \frac{(v \sin \theta)^2}{2g} \]

4. **Independence of Motion:**
- The horizontal motion is uniform, as there is no acceleration in the horizontal direction (ignoring air resistance). The horizontal distance traveled can be calculated as:
\[ S_x = v_x \cdot T = (v \cos \theta) \cdot T \]
- The vertical and horizontal motions can be treated independently because the horizontal motion does not affect the vertical motion and vice versa.

### Conclusion:
- **Statement-1** is true: The time of flight \(T = \frac{2v \sin \theta}{g}\) is correct.
- **Statement-2** is also true: The vertical and horizontal motions can indeed be treated independently.