A body is thrown from the surface of the earth with velocity `(gR_(e))//12`, where `R_(e)` is the radius of the earth at some angle from the vertical. If the maximum height reached by the body is `R_(e)//4`, then the angle of projection with the vertical is

Options

A. `sin^(-1)((sqrt(5))/4)`
B. `cos^(-1)((sqrt(5))/4)`
C. `sin^(-1)((sqrt(3))/2)`
D. none of these

Explanation

To solve the problem, we need to find the angle of projection of a body thrown from the surface of the Earth with a specific initial velocity and reaching a certain maximum height.

### Given:
- Initial velocity \( v_0 = \frac{g R_e}{12} \)
- Maximum height \( h = \frac{R_e}{4} \)
- Acceleration due to gravity \( g \)

### Step 1: Use the formula for maximum height in projectile motion
The maximum height \( h \) reached by a projectile is given by the formula:
\[ h = \frac{v_y^2}{2g} \]
where \( v_y \) is the vertical component of the initial velocity.

### Step 2: Find the vertical component of the initial velocity
Let \( \theta \) be the angle of projection with the vertical. The vertical component of the initial velocity can be expressed as:
\[ v_y = v_0 \cos(\theta) \]
Substituting the value of \( v_0 \):
\[ v_y = \frac{g R_e}{12} \cos(\theta) \]

### Step 3: Substitute \( v_y \) into the height formula
Now, substituting \( v_y \) into the height formula:
\[ h = \frac{\left(\frac{g R_e}{12} \cos(\theta)\right)^2}{2g} \]
This simplifies to:
\[ h = \frac{g^2 R_e^2 \cos^2(\theta)}{288g} = \frac{g R_e^2 \cos^2(\theta)}{288} \]

### Step 4: Set the maximum height equal to the given height
We know that \( h = \frac{R_e}{4} \). Setting the two expressions for height equal gives:
\[ \frac{g R_e^2 \cos^2(\theta)}{288} = \frac{R_e}{4} \]

### Step 5: Simplify the equation
To eliminate \( R_e \) from both sides, we can multiply both sides by \( 288 \) and divide by \( g \):
\[ R_e \cos^2(\theta) = 72 \]

### Step 6: Solve for \( \cos^2(\theta) \)
Rearranging gives:
\[ \cos^2(\theta) = \frac{72}{R_e} \]

### Step 7: Find the angle \( \theta \)
To find \( \theta \), we take the cosine inverse:
\[ \theta = \cos^{-1}\left(\sqrt{\frac{72}{R_e}}\right) \]

### Conclusion
Thus, the angle of projection with the vertical is:
\[ \theta = \cos^{-1}\left(\sqrt{\frac{72}{R_e}}\right) \]