A body is projected horizontally from the surface of the earth (radius `= R`) with a velocity equal to `n` times the escape velocity. Neglect rotational effect of the earth. The maximum height attained by the body from the earth `s` surface is `R//2`. Then, `n` must be

Options

A. `(a)sqrt(0.6)`
B. `(b)(sqrt(3))//2`
C. `(c)sqrt(0.4)`
D. `(d)1//2`

Explanation

To solve the problem, we need to find the value of `n` when a body is projected horizontally from the surface of the Earth with a velocity equal to `n` times the escape velocity, and it reaches a maximum height of `R/2` above the Earth's surface.

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
The escape velocity (Ve) from the surface of the Earth is given by:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
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 body is projected with a velocity \( V_1 = nV_e \). Therefore, substituting for \( V_e \):
\[ V_1 = n \sqrt{\frac{2GM}{R}} \]

3. **Conservation of Angular Momentum**:
The initial angular momentum \( L_i \) about the center of the Earth is:
\[ L_i = mV_1R = mnV_eR \]
At the maximum height \( h = \frac{R}{2} \), the distance from the center of the Earth becomes \( R + \frac{R}{2} = \frac{3R}{2} \). The final angular momentum \( L_f \) is:
\[ L_f = mV_2 \left(\frac{3R}{2}\right) \]
By conservation of angular momentum, we have:
\[ mnV_eR = mV_2 \left(\frac{3R}{2}\right) \]
Canceling \( m \) and \( R \) gives:
\[ nV_e = \frac{3}{2} V_2 \]
Thus, we can express \( V_2 \):
\[ V_2 = \frac{2}{3}nV_e \]

4. **Conservation of Energy**:
The initial mechanical energy \( E_i \) is the sum of kinetic and potential energy:
\[ E_i = \frac{1}{2} m V_1^2 - \frac{GMm}{R} \]
Substituting \( V_1 \):
\[ E_i = \frac{1}{2} m (nV_e)^2 - \frac{GMm}{R} \]
The final mechanical energy \( E_f \) at height \( \frac{R}{2} \) is:
\[ E_f = \frac{1}{2} m V_2^2 - \frac{GMm}{\frac{3R}{2}} \]
Substituting \( V_2 \):
\[ E_f = \frac{1}{2} m \left(\frac{2}{3}nV_e\right)^2 - \frac{2GMm}{3R} \]

5. **Setting Initial Energy Equal to Final Energy**:
Setting \( E_i = E_f \):
\[ \frac{1}{2} m n^2 \frac{2GM}{R} - \frac{GMm}{R} = \frac{2}{9} m n^2 \frac{2GM}{R} - \frac{2GMm}{3R} \]
Cancel \( m \) and \( GM \):
\[ \frac{n^2}{R} - \frac{1}{R} = \frac{4n^2}{9R} - \frac{2}{3R} \]
Multiplying through by \( R \):
\[ n^2 - 1 = \frac{4n^2}{9} - \frac{2}{3} \]

6. **Solving for n**:
Rearranging gives:
\[ 9(n^2 - 1) = 4n^2 - 6 \]
\[ 9n^2 - 9 = 4n^2 - 6 \]
\[ 5n^2 = 3 \]
\[ n^2 = \frac{3}{5} \]
\[ n = \sqrt{\frac{3}{5}} = \frac{\sqrt{15}}{5} \]

### Final Answer:
Thus, the value of \( n \) is:
\[ n = \sqrt{\frac{3}{5}} \approx 0.7746 \]