A particle is projected horizontally with speed `(20)/(sqrt(3))m//s`, from some height at t=0. At what time its velocity will make `60^(@)` angle with initial velocity ?

Options

A. 1 sec
B. 2 sec
C. `1.5` sec
D. `2.5` sec

Answer

To solve the problem, we need to determine the time at which the velocity of a horizontally projected particle makes a 60-degree angle with its initial velocity.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The particle is projected horizontally with an initial speed \( u = \frac{20}{\sqrt{3}} \, \text{m/s} \).
- The initial vertical velocity \( u_y = 0 \, \text{m/s} \) since the projection is horizontal.

2. **Understanding the Velocity Components**:
- The horizontal component of the velocity \( V_x \) remains constant throughout the motion:
\[ V_x = u = \frac{20}{\sqrt{3}} \, \text{m/s} \]
- The vertical component of the velocity \( V_y \) at time \( t \) can be calculated using the formula:
\[ V_y = g t \]
where \( g \) is the acceleration due to gravity, approximately \( 10 \, \text{m/s}^2 \).

3. **Setting Up the Angle Condition**:
- We need to find the time \( t \) when the resultant velocity \( V \) makes a \( 60^\circ \) angle with the horizontal. The angle condition can be expressed using the tangent function:
\[ \tan(60^\circ) = \frac{V_y}{V_x} \]
- Since \( \tan(60^\circ) = \sqrt{3} \), we have:
\[ \sqrt{3} = \frac{V_y}{V_x} \]

4. **Substituting the Velocity Components**:
- Substitute \( V_y \) and \( V_x \):
\[ \sqrt{3} = \frac{g t}{\frac{20}{\sqrt{3}}} \]
- Rearranging gives:
\[ \sqrt{3} \cdot \frac{20}{\sqrt{3}} = g t \]
- Simplifying:
\[ 20 = 10 t \]

5. **Solving for Time**:
- From the equation \( 20 = 10 t \):
\[ t = \frac{20}{10} = 2 \, \text{seconds} \]

### Final Answer:
The time at which the velocity of the particle makes a \( 60^\circ \) angle with the initial velocity is \( t = 2 \, \text{seconds} \).

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