An object is projected from the earth's surface with escape velocity at `30^(@)` with horizontal. What is the angle made by the velocity with horizontal when the object reaches a height 2R from the earth's surface ? R is the radius of the earth. Horizontal can be considered as a line parallel to the tangent at the earth's surface just below the object .

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `15^(@)`

Answer

To solve the problem, we need to determine the angle made by the velocity of an object with the horizontal when it reaches a height of \(2R\) from the Earth's surface, given that it was projected with an escape velocity at an angle of \(30^\circ\) with the horizontal.

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
The escape velocity \(V_E\) from the Earth's surface is given by the formula:
\[ V_E = \sqrt{\frac{2GM}{R}} \]
where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(R\) is the radius of the Earth.

2. **Initial Conditions**:
The object is projected with an escape velocity at an angle of \(30^\circ\). The initial velocity can be broken down into horizontal and vertical components:
\[ V_{x0} = V_E \cos(30^\circ) = V_E \cdot \frac{\sqrt{3}}{2} \]
\[ V_{y0} = V_E \sin(30^\circ) = V_E \cdot \frac{1}{2} \]

3. **Using Conservation of Energy**:
We can apply the conservation of mechanical energy to find the velocity of the object when it reaches the height \(2R\).
- At the initial point (height = 0):
\[ E_1 = K.E_1 + P.E_1 = \frac{1}{2} m V_E^2 - \frac{GMm}{R} \]
- At the height \(h = 2R\):
\[ E_2 = K.E_2 + P.E_2 = \frac{1}{2} m V^2 - \frac{GMm}{3R} \]

Setting \(E_1 = E_2\):
\[ \frac{1}{2} m V_E^2 - \frac{GMm}{R} = \frac{1}{2} m V^2 - \frac{GMm}{3R} \]

4. **Simplifying the Energy Equation**:
Cancel \(m\) from both sides and rearranging gives:
\[ \frac{1}{2} V_E^2 - \frac{GM}{R} = \frac{1}{2} V^2 - \frac{GM}{3R} \]
Rearranging further, we find:
\[ \frac{1}{2} V^2 = \frac{1}{2} V_E^2 - \frac{GM}{R} + \frac{GM}{3R} \]
\[ \frac{1}{2} V^2 = \frac{1}{2} V_E^2 - \frac{2GM}{3R} \]

5. **Finding the Final Velocity**:
Substitute \(V_E\) into the equation:
\[ V^2 = V_E^2 - \frac{4GM}{3R} \]
Using \(V_E^2 = \frac{2GM}{R}\):
\[ V^2 = \frac{2GM}{R} - \frac{4GM}{3R} = \frac{6GM}{3R} - \frac{4GM}{3R} = \frac{2GM}{3R} \]
Thus,
\[ V = V_E \cdot \frac{1}{\sqrt{3}} \]

6. **Angular Momentum Conservation**:
The angular momentum is conserved. The initial angular momentum \(L_1\) and final angular momentum \(L_2\) are:
\[ L_1 = m V_E \cos(30^\circ) R \]
\[ L_2 = m V \cos(\alpha) (3R) \]

Setting \(L_1 = L_2\):
\[ V_E \cdot \frac{\sqrt{3}}{2} R = V \cos(\alpha) (3R) \]

7. **Substituting for \(V\)**:
Substitute \(V = \frac{V_E}{\sqrt{3}}\):
\[ V_E \cdot \frac{\sqrt{3}}{2} R = \frac{V_E}{\sqrt{3}} \cos(\alpha) (3R) \]
Cancel \(V_E\) and \(R\):
\[ \frac{\sqrt{3}}{2} = \frac{3 \cos(\alpha)}{\sqrt{3}} \]
Rearranging gives:
\[ \cos(\alpha) = \frac{1}{2} \]

8. **Finding the Angle**:
The angle \(\alpha\) corresponding to \(\cos(\alpha) = \frac{1}{2}\) is:
\[ \alpha = 60^\circ \]

### Final Answer:
The angle made by the velocity with the horizontal when the object reaches a height \(2R\) from the Earth's surface is \(60^\circ\).