A ball is projected with a velocity `20 sqrt(3) ms^(-1)` at angle `60^(@)` to the horizontal. The time interval after which the velocity vector will make an angle `30^(@)` to the horizontal is (Take, `g = 10 ms^(-2))`

Options

A. 5s
B. 2s
C. 1s
D. 3s

Explanation

To solve the problem, we need to find the time interval after which the velocity vector of the ball makes an angle of 30 degrees with the horizontal. Let's break it down step by step.

### Step 1: Determine the initial velocity components
The initial velocity \( V_0 \) is given as \( 20\sqrt{3} \, \text{m/s} \) at an angle of \( 60^\circ \).

1. **Horizontal component \( V_{x0} \)**:
\[ V_{x0} = V_0 \cos(60^\circ) = 20\sqrt{3} \cdot \frac{1}{2} = 10\sqrt{3} \, \text{m/s} \]

2. **Vertical component \( V_{y0} \)**:
\[ V_{y0} = V_0 \sin(60^\circ) = 20\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 30 \, \text{m/s} \]

### Step 2: Write the equations for velocity at time \( t \)
The horizontal velocity \( V_x \) remains constant since there is no horizontal acceleration:
\[ V_x = 10\sqrt{3} \, \text{m/s} \]

The vertical velocity \( V_y \) changes due to gravity:
\[ V_y = V_{y0} - g t = 30 - 10t \, \text{m/s} \]

### Step 3: Set up the condition for the angle
The velocity vector makes an angle of \( 30^\circ \) with the horizontal, which means:
\[ \tan(30^\circ) = \frac{V_y}{V_x} \]
From trigonometric values, we know:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

### Step 4: Substitute the values into the equation
Substituting the expressions for \( V_y \) and \( V_x \):
\[ \frac{30 - 10t}{10\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Step 5: Solve for \( t \)
Cross-multiplying gives:
\[ 30 - 10t = 10 \]
Rearranging the equation:
\[ 30 - 10 = 10t \implies 20 = 10t \implies t = 2 \, \text{seconds} \]

### Final Answer
The time interval after which the velocity vector makes an angle of \( 30^\circ \) with the horizontal is \( t = 2 \, \text{seconds} \).

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