A projectile is thrown in the upward direction making an angle of `60^@` with the horizontal direction with a velocity of `150 ms^-1`. Then the time after which its inclination with the horizontal is `45^@` is

Options

A. `15 (sqrt(3) - 1) s`
B. `15 (sqrt(3) + 1) s`
C. `7.5 (sqrt(3) - 1) s`
D. `7.5 (sqrt(3) + 1) s`

Explanation

To solve the problem, we need to determine the time at which the projectile's velocity makes an angle of 45° with the horizontal after being thrown at an angle of 60° with an initial velocity of 150 m/s.

### Step-by-Step Solution:

1. **Identify the Components of the Initial Velocity:**
- The initial velocity \( u = 150 \, \text{m/s} \).
- The angle of projection \( \theta = 60^\circ \).
- The horizontal component of the initial velocity \( u_x \) is given by:
\[ u_x = u \cos(\theta) = 150 \cos(60^\circ) = 150 \times \frac{1}{2} = 75 \, \text{m/s} \]
- The vertical component of the initial velocity \( u_y \) is given by:
\[ u_y = u \sin(\theta) = 150 \sin(60^\circ) = 150 \times \frac{\sqrt{3}}{2} = 75\sqrt{3} \, \text{m/s} \]

2. **Determine the Condition for 45° Inclination:**
- At the time \( t \) when the projectile's velocity makes an angle of 45° with the horizontal, the horizontal and vertical components of the velocity must be equal:
\[ v_y = v_x \]
- The horizontal component of the velocity \( v_x \) remains constant:
\[ v_x = u_x = 75 \, \text{m/s} \]
- The vertical component of the velocity \( v_y \) at time \( t \) is given by:
\[ v_y = u_y - g t = 75\sqrt{3} - 10t \]

3. **Set the Components Equal:**
- Setting the vertical and horizontal components equal gives:
\[ 75\sqrt{3} - 10t = 75 \]

4. **Solve for Time \( t \):**
- Rearranging the equation:
\[ 75\sqrt{3} - 75 = 10t \]
- Simplifying:
\[ 75(\sqrt{3} - 1) = 10t \]
- Dividing both sides by 10:
\[ t = \frac{75(\sqrt{3} - 1)}{10} = 7.5(\sqrt{3} - 1) \, \text{s} \]

### Final Answer:
The time after which the projectile's inclination with the horizontal is \( 45^\circ \) is:
\[ t = 7.5(\sqrt{3} - 1) \, \text{s} \]