A particle is projected with velocity 50 m/s at an angle ` 60^(@)` with the horizontal from the ground. The time after which its velocity will make an angle ` 45^(@)` with the horizontal is

Options

A. 2.5 s
B. 1.83 s
C. 2.37 s
D. 3.72 m

Answer

To solve the problem step by step, we need to analyze the motion of the particle projected at a velocity of 50 m/s at an angle of 60 degrees with the horizontal. We want to find the time at which the velocity of the particle makes an angle of 45 degrees with the horizontal.

### Step 1: Break down the initial velocity into components
The initial velocity \( u = 50 \, \text{m/s} \) can be resolved into horizontal and vertical components using trigonometric functions.

- Horizontal component \( u_x = u \cos(60^\circ) = 50 \cdot \frac{1}{2} = 25 \, \text{m/s} \)
- Vertical component \( u_y = u \sin(60^\circ) = 50 \cdot \frac{\sqrt{3}}{2} = 25\sqrt{3} \, \text{m/s} \)

### Step 2: Determine the conditions for the velocity to make a 45-degree angle
For the velocity to make an angle of 45 degrees with the horizontal, the vertical component of the velocity \( v_y \) must equal the horizontal component \( v_x \).

### Step 3: Express the horizontal and vertical components of velocity at time \( t \)
Since there is no acceleration in the horizontal direction, the horizontal component remains constant:
- \( v_x = u_x = 25 \, \text{m/s} \)

The vertical component of velocity changes due to gravity:
- \( v_y = u_y - g t \)
- \( v_y = 25\sqrt{3} - 10t \)

### Step 4: Set up the equation for the condition \( v_y = v_x \)
We set the vertical component equal to the horizontal component:
\[ 25\sqrt{3} - 10t = 25 \]

### Step 5: Solve for \( t \)
Rearranging the equation gives:
\[ 25\sqrt{3} - 25 = 10t \]
\[ 10t = 25(\sqrt{3} - 1) \]
\[ t = \frac{25(\sqrt{3} - 1)}{10} \]
\[ t = 2.5(\sqrt{3} - 1) \]

Calculating \( \sqrt{3} \approx 1.732 \):
\[ t = 2.5(1.732 - 1) \]
\[ t = 2.5(0.732) \]
\[ t \approx 1.83 \, \text{s} \]

### Step 6: Check for the second instance when the angle is 45 degrees
The particle will also make a 45-degree angle on its way down. The vertical component will be negative:
\[ -v_y = u_y - g t \]
Setting this equal to \( v_x \):
\[ - (25\sqrt{3} - 10t) = 25 \]
This leads to:
\[ 10t = 25\sqrt{3} + 25 \]
\[ t = \frac{25(\sqrt{3} + 1)}{10} \]
\[ t = 2.5(\sqrt{3} + 1) \]
Calculating this gives:
\[ t \approx 2.5(1.732 + 1) \]
\[ t \approx 2.5(2.732) \]
\[ t \approx 6.83 \, \text{s} \]

### Final Answer
The times at which the velocity of the particle makes a 45-degree angle with the horizontal are approximately \( 1.83 \, \text{s} \) and \( 6.83 \, \text{s} \).