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 will analyze the motion of the particle and calculate the time at which its velocity makes an angle of 45 degrees with the horizontal.

### Step 1: Determine the initial velocity components
The initial velocity \( V_0 \) is given as 50 m/s at an angle of 60 degrees with the horizontal. We can break this velocity into its horizontal and vertical components.

- Horizontal component \( V_{0x} = V_0 \cos(60^\circ) = 50 \cdot \frac{1}{2} = 25 \, \text{m/s} \)
- Vertical component \( V_{0y} = V_0 \sin(60^\circ) = 50 \cdot \frac{\sqrt{3}}{2} = 25\sqrt{3} \, \text{m/s} \)

### Step 2: Understand the condition for the angle of 45 degrees
At the time \( t \) when the velocity makes an angle of 45 degrees with the horizontal, the horizontal and vertical components of the velocity will be equal. Thus, we need to find the time \( t \) when:

\[ V_x = V_y \]

### Step 3: Write the expressions for the velocity components at time \( t \)
The horizontal component of velocity remains constant (since there is no horizontal acceleration):

- \( V_x = V_{0x} = 25 \, \text{m/s} \)

The vertical component of velocity changes due to gravity:

- \( V_y = V_{0y} - g t = 25\sqrt{3} - 9.8 t \)

### Step 4: Set the horizontal and vertical components equal
To find the time when the angle is 45 degrees, we set the horizontal and vertical components equal:

\[ V_{0x} = V_y \]

Substituting the expressions we have:

\[ 25 = 25\sqrt{3} - 9.8 t \]

### Step 5: Solve for time \( t \)
Rearranging the equation gives:

\[ 9.8 t = 25\sqrt{3} - 25 \]

Now, we can factor out 25:

\[ 9.8 t = 25(\sqrt{3} - 1) \]

Now, divide both sides by 9.8:

\[ t = \frac{25(\sqrt{3} - 1)}{9.8} \]

### Step 6: Calculate the numerical value
Using \( \sqrt{3} \approx 1.732 \):

\[ t = \frac{25(1.732 - 1)}{9.8} \]
\[ t = \frac{25(0.732)}{9.8} \]
\[ t \approx \frac{18.3}{9.8} \approx 1.87 \, \text{s} \]

### Final Answer
The time after which the velocity will make an angle of 45 degrees with the horizontal is approximately **1.87 seconds**.

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