A particle is projected with initial velocity u making an angle `alpha` with the horizontal, its time of flight will be given by

Options

A. `(2usinalpha)/(g)`
B. `(2u^(2)sinalpha)/(g)`
C. `(usinalpha)/(g)`
D. `(u^(2)sinalpha)/(g)`

Answer

To find the time of flight of a particle projected with an initial velocity \( u \) at an angle \( \alpha \) with the horizontal, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The particle is projected at an angle \( \alpha \) with an initial velocity \( u \). The motion can be analyzed in two dimensions: horizontal (x-axis) and vertical (y-axis).

2. **Components of Initial Velocity**:
- The initial velocity can be broken down into its horizontal and vertical components:
- \( u_x = u \cos \alpha \) (horizontal component)
- \( u_y = u \sin \alpha \) (vertical component)

3. **Time of Flight Definition**:
- The time of flight is the total time the particle is in the air before it returns to the ground (displacement in the y-direction is zero).

4. **Using the Vertical Motion Equation**:
- The vertical motion can be described using the equation of motion:
\[ s_y = u_y t + \frac{1}{2} a_y t^2 \]
- Here, \( s_y \) is the vertical displacement, \( u_y \) is the initial vertical velocity, \( a_y \) is the vertical acceleration (which is \( -g \), where \( g \) is the acceleration due to gravity), and \( t \) is the time of flight.

5. **Setting Up the Equation**:
- Since the particle returns to the ground, the vertical displacement \( s_y = 0 \):
\[ 0 = (u \sin \alpha) t + \frac{1}{2} (-g) t^2 \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ 0 = u \sin \alpha \cdot t - \frac{1}{2} g t^2 \]
- Factoring out \( t \):
\[ t \left( u \sin \alpha - \frac{1}{2} g t \right) = 0 \]

7. **Finding the Non-Zero Solution**:
- The solutions to this equation are \( t = 0 \) (at the moment of projection) and:
\[ u \sin \alpha - \frac{1}{2} g t = 0 \]
- Solving for \( t \):
\[ u \sin \alpha = \frac{1}{2} g t \]
\[ t = \frac{2 u \sin \alpha}{g} \]

8. **Final Result**:
- Thus, the time of flight \( T \) is given by:
\[ T = \frac{2 u \sin \alpha}{g} \]