A particle at a height `h` from the ground is projected with an angle `30^@` from the horizontal, it strikes the ground making angle `45^@` with horizontal. It is again projected from the same point with the same speed but with an angle of `60^@` with horizontal. Find the angle it makes with the horizontal when it strikes the ground

Options

A. `tan^-1(4)`
B. `tan^-1(5)`
C. `tan^-1(sqrt5)`
D. `tan^-1(sqrt3)`

Explanation

To solve the problem step by step, we need to analyze the projectile motion of the particle in two different scenarios: first when it is projected at an angle of \(30^\circ\) and then at \(60^\circ\).

### Step 1: Analyze the first projection at \(30^\circ\)

1. **Components of Initial Velocity**:
Let the initial speed of projection be \(u\).
- Horizontal component:
\[ u_x = u \cos(30^\circ) = u \cdot \frac{\sqrt{3}}{2} \]
- Vertical component:
\[ u_y = u \sin(30^\circ) = u \cdot \frac{1}{2} \]