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 find the time after which the projectile's inclination with the horizontal is \( 45^\circ \). We start by breaking down the initial velocity into its horizontal and vertical components.

### Step 1: Break down the initial velocity
The initial velocity \( u = 150 \, \text{m/s} \) is given at an angle of \( 60^\circ \) to the horizontal. We can find the horizontal and vertical components of the initial velocity using trigonometric functions.

- Horizontal component \( u_x = u \cos(60^\circ) = 150 \cdot \frac{1}{2} = 75 \, \text{m/s} \)
- Vertical component \( u_y = u \sin(60^\circ) = 150 \cdot \frac{\sqrt{3}}{2} = 75\sqrt{3} \, \text{m/s} \)

### Step 2: Determine the velocity at \( 45^\circ \)
At the time when the projectile's inclination is \( 45^\circ \), the horizontal and vertical components of the velocity will be equal. Let \( v \) be the magnitude of the velocity at that time.

From the geometry of the situation:
- Horizontal component at \( 45^\circ \): \( v_x = v \cos(45^\circ) = \frac{v}{\sqrt{2}} \)
- Vertical component at \( 45^\circ \): \( v_y = v \sin(45^\circ) = \frac{v}{\sqrt{2}} \)

Since the horizontal component of the velocity remains constant:
\[ \frac{v}{\sqrt{2}} = 75 \implies v = 75\sqrt{2} \, \text{m/s} \]

### Step 3: Find the vertical component of the velocity
Using the relationship established, we can find the vertical component of the velocity at the time when the inclination is \( 45^\circ \):
\[ v_y = \frac{v}{\sqrt{2}} = \frac{75\sqrt{2}}{\sqrt{2}} = 75 \, \text{m/s} \]

### Step 4: Apply the equations of motion
Now, we can use the first equation of motion in the vertical direction to find the time \( t \):
\[ v_y = u_y - g t \]
Substituting the known values:
\[ 75 = 75\sqrt{3} - g t \]
Where \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \).

Rearranging gives:
\[ g t = 75\sqrt{3} - 75 \]
\[ t = \frac{75(\sqrt{3} - 1)}{g} \]

### Step 5: Substitute the value of \( g \)
Using \( g \approx 9.8 \, \text{m/s}^2 \):
\[ t = \frac{75(\sqrt{3} - 1)}{9.8} \]

### Final Result
Calculating this gives the time after which the projectile's inclination with the horizontal is \( 45^\circ \).