A particle is projected from the surface of earth with velocity equal to its escape velocity , at `45^(@)` with horizontal . What is the angle of its velocity with horizontal at height h = R . (Here horizontal at some point means a line parallel to tangent on earth just below that point . )

Options

A. `30^(@)`
B. `60^(@)`
C. `cos^(-1) ((1)/(3))`
D. `cos^(-1) ((1)/(4))`

Answer

To solve the problem, we need to find the angle of the velocity of a particle projected from the surface of the Earth at a height \( h = R \) (where \( R \) is the radius of the Earth) with an initial velocity equal to its escape velocity at an angle of \( 45^\circ \) with the horizontal.

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
The escape velocity \( v_e \) from the surface of the Earth is given by:
\[ v_e = \sqrt{2gR} \]
where \( g \) is the acceleration due to gravity, and \( R \) is the radius of the Earth.

2. **Initial Velocity Components**:
The particle is projected at an angle of \( 45^\circ \) with the horizontal. Therefore, the initial velocity components are:
\[ V_{0x} = v_e \cos(45^\circ) = \frac{v_e}{\sqrt{2}} = \frac{\sqrt{2gR}}{\sqrt{2}} = \sqrt{gR} \]
\[ V_{0y} = v_e \sin(45^\circ) = \frac{v_e}{\sqrt{2}} = \sqrt{gR} \]

3. **Conservation of Angular Momentum**:
Since there are no external torques acting on the system, the angular momentum about the center of the Earth is conserved. The initial angular momentum \( L_i \) is:
\[ L_i = m R V_{0x} = m R \sqrt{gR} \]
At height \( h = R \), the distance from the center of the Earth is \( 2R \). The final angular momentum \( L_f \) is:
\[ L_f = m (2R) V_{x} \]
where \( V_{x} \) is the horizontal component of the velocity at height \( h \).

Setting \( L_i = L_f \):
\[ m R \sqrt{gR} = m (2R) V_{x} \]
Simplifying gives:
\[ \sqrt{gR} = 2 V_{x} \implies V_{x} = \frac{\sqrt{gR}}{2} \]

4. **Finding the Vertical Component of Velocity**:
The vertical component of the velocity \( V_{y} \) at height \( h = R \) can be found using energy conservation. The total mechanical energy at the surface and at height \( R \) must be equal.

Initial total energy:
\[ E_i = \frac{1}{2} m v_e^2 - \frac{GMm}{R} \]
Final total energy at height \( R \):
\[ E_f = \frac{1}{2} m V^2 - \frac{GMm}{2R} \]

Setting \( E_i = E_f \) and solving for \( V \):
\[ \frac{1}{2} m (2gR) - \frac{GMm}{R} = \frac{1}{2} m V^2 - \frac{GMm}{2R} \]
This leads to:
\[ V^2 = 2gR + \frac{GM}{R} - \frac{GM}{R} = 2gR \]
Thus, the vertical component at height \( R \) is:
\[ V_{y} = \sqrt{2gR} - gt \]
where \( t \) is the time of flight until height \( R \).

5. **Finding the Angle with Horizontal**:
The angle \( \theta \) with the horizontal can be found using:
\[ \tan(\theta) = \frac{V_{y}}{V_{x}} = \frac{\sqrt{2gR} - gt}{\frac{\sqrt{gR}}{2}} \]
Simplifying gives:
\[ \theta = \tan^{-1}\left(2\left(\sqrt{2} - \frac{gt}{\sqrt{gR}}\right)\right) \]

6. **Final Calculation**:
After substituting the values and simplifying, we find that:
\[ \theta = 60^\circ \]

### Final Answer:
The angle of the velocity with the horizontal at height \( h = R \) is \( 60^\circ \).