The total speed of a projectile at its greater height is `sqrt(6/7)` of its speed when it is at half of its greatest height. The angle of projection will be :-

Options

A. `60^(@)`
B. `45^(@)`
C. `30^(@)`
D. `50^(@)`

Explanation

To solve the problem, we need to find the angle of projection \( \theta \) given the relationship between the speeds of a projectile at its greatest height and at half of its greatest height.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The speed at the greatest height is given as \( v_{max} = \sqrt{\frac{6}{7}} v_{half} \).
- At the greatest height, the vertical component of the velocity is zero, and the horizontal component remains \( v_0 \cos \theta \).

2. **Velocity at Greatest Height**:
- At the greatest height, the total speed \( v_{max} \) is equal to the horizontal component:
\[ v_{max} = v_0 \cos \theta \]

3. **Velocity at Half of the Greatest Height**:
- The height \( h \) of the projectile is given by:
\[ h = \frac{v_0^2 \sin^2 \theta}{2g} \]
- Therefore, half of the greatest height is \( \frac{h}{2} = \frac{v_0^2 \sin^2 \theta}{4g} \).
- The vertical component of the velocity at half height can be found using the kinematic equation:
\[ v_y^2 = u_y^2 + 2as \]
where \( u_y = v_0 \sin \theta \), \( a = -g \), and \( s = \frac{h}{2} \):
\[ v_y^2 = (v_0 \sin \theta)^2 - 2g \left(\frac{v_0^2 \sin^2 \theta}{4g}\right) \]
Simplifying this gives:
\[ v_y^2 = v_0^2 \sin^2 \theta - \frac{v_0^2 \sin^2 \theta}{2} = \frac{v_0^2 \sin^2 \theta}{2} \]
Thus,
\[ v_y = \frac{v_0 \sin \theta}{\sqrt{2}} \]

4. **Total Velocity at Half Height**:
- The total velocity \( v_{half} \) at half height is:
\[ v_{half} = \sqrt{v_x^2 + v_y^2} = \sqrt{(v_0 \cos \theta)^2 + \left(\frac{v_0 \sin \theta}{\sqrt{2}}\right)^2} \]
\[ v_{half} = \sqrt{v_0^2 \cos^2 \theta + \frac{v_0^2 \sin^2 \theta}{2}} = v_0 \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}} \]

5. **Setting Up the Equation**:
- From the problem statement, we know:
\[ v_{max} = \sqrt{\frac{6}{7}} v_{half} \]
- Substituting the expressions for \( v_{max} \) and \( v_{half} \):
\[ v_0 \cos \theta = \sqrt{\frac{6}{7}} \cdot v_0 \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}} \]
- Dividing both sides by \( v_0 \) (assuming \( v_0 \neq 0 \)):
\[ \cos \theta = \sqrt{\frac{6}{7}} \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}} \]

6. **Squaring Both Sides**:
- Squaring both sides gives:
\[ \cos^2 \theta = \frac{6}{7} \left(\cos^2 \theta + \frac{\sin^2 \theta}{2}\right) \]
- Expanding and rearranging:
\[ 7 \cos^2 \theta = 6 \cos^2 \theta + 3 \sin^2 \theta \]
\[ \cos^2 \theta = 3 \sin^2 \theta \]

7. **Using the Identity**:
- Using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \cos^2 \theta = 3(1 - \cos^2 \theta) \]
\[ 4 \cos^2 \theta = 3 \]
\[ \cos^2 \theta = \frac{3}{4} \implies \sin^2 \theta = \frac{1}{4} \]

8. **Finding the Angle**:
- Thus, \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{1/4}{3/4} = \frac{1}{3} \).
- Therefore, \( \tan \theta = \frac{1}{\sqrt{3}} \) which gives:
\[ \theta = 30^\circ \]

### Final Answer:
The angle of projection \( \theta \) is \( 30^\circ \).