When a body is thrown with a velocity u making an angle `theta` with the horizontal plane, the maximum distance covered by it in horizontal direction is

Options

A. `(u^(2) sin theta)/(g)`
B. `(u^(2) sin 2theta)/(2g)`
C. `(u^(2)sin 2theta)/(g)`
D. `(u^(2) cos 2 theta)/(g)`

Answer

To find the maximum distance covered by a body thrown with an initial velocity \( u \) at an angle \( \theta \) with the horizontal plane, we can follow these steps:

### Step 1: Understand the Components of Velocity
When the body is thrown, its initial velocity \( u \) can be resolved into two components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

### Step 2: Determine the Time of Flight
The time of flight \( T \) for a projectile can be calculated using the vertical motion. The body will rise to a maximum height and then fall back to the ground. The time to reach the maximum height is given by:
\[ t_{up} = \frac{u_y}{g} = \frac{u \sin \theta}{g} \]
Since the time to go up is equal to the time to come down, the total time of flight \( T \) is:
\[ T = 2 t_{up} = 2 \frac{u \sin \theta}{g} \]

### Step 3: Calculate the Horizontal Range
The horizontal range \( R \) is the horizontal distance covered during the total time of flight. Since there is no horizontal acceleration, the range can be calculated as:
\[ R = u_x \cdot T = (u \cos \theta) \cdot \left(2 \frac{u \sin \theta}{g}\right) \]
Substituting the values:
\[ R = 2 \frac{u \cos \theta \cdot u \sin \theta}{g} \]

### Step 4: Simplify the Expression
Using the trigonometric identity \( \sin(2\theta) = 2 \sin \theta \cos \theta \), we can simplify the expression for the range:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]

### Final Result
Thus, the maximum distance covered by the body in the horizontal direction is:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]