The time of fight of an object projected with speed ` 20 ms^(-1)` at an angle ` 30^(@)` with the horizontal , is

Options

A. 1 s
B. 4 s
C. 2 s
D. 6s

Answer

To find the time of flight of an object projected with a speed of \( 20 \, \text{m/s} \) at an angle of \( 30^\circ \) with the horizontal, we can follow these steps:

### Step 1: Identify the components of the initial velocity
The initial velocity \( u \) can be broken down into two components:
- Horizontal component \( u_x = u \cos \theta \)
- Vertical component \( u_y = u \sin \theta \)

Given:
- \( u = 20 \, \text{m/s} \)
- \( \theta = 30^\circ \)

Calculating the vertical component:
\[ u_y = 20 \sin(30^\circ) = 20 \times \frac{1}{2} = 10 \, \text{m/s} \]

### Step 2: Use the formula for time of flight
The time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2u_y}{g} \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 3: Substitute the values into the formula
Substituting \( u_y \) and \( g \) into the formula:
\[ T = \frac{2 \times 10}{10} = \frac{20}{10} = 2 \, \text{s} \]

### Conclusion
The time of flight of the object is \( 2 \, \text{seconds} \).

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