A particle is projected from the ground with velocity u making an angle `theta` with the horizontal. At half of its maximum heights,

Options

A. its horizontal velocity is `ucostheta`
B. its vertical velocity is `(usintheta)/(sqrt2)`
C. its velocity is `((1+Cos^(2)theta)/(2))^(1/2)`
D. all the above are true

Answer

To solve the problem, we need to find the velocity of a particle at half of its maximum height when it is projected from the ground with an initial velocity \( u \) at an angle \( \theta \) with the horizontal.

### Step-by-step Solution:

1. **Determine Maximum Height**:
The maximum height \( H \) reached by the projectile can be calculated using the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

2. **Calculate Half of Maximum Height**:
Half of the maximum height \( h \) is:
\[ h = \frac{H}{2} = \frac{u^2 \sin^2 \theta}{4g} \]

3. **Find Vertical Velocity at Half Maximum Height**:
The vertical component of the velocity \( v_y \) at half of the maximum height can be found using the kinematic equation:
\[ v_y^2 = u^2 \sin^2 \theta - 2g h \]
Substituting \( h \):
\[ v_y^2 = u^2 \sin^2 \theta - 2g \left(\frac{u^2 \sin^2 \theta}{4g}\right) \]
Simplifying this gives:
\[ v_y^2 = u^2 \sin^2 \theta - \frac{u^2 \sin^2 \theta}{2} = \frac{u^2 \sin^2 \theta}{2} \]
Therefore, the vertical component of the velocity is:
\[ v_y = \frac{u \sin \theta}{\sqrt{2}} \]

4. **Calculate Horizontal Velocity**:
The horizontal component of the velocity \( v_x \) remains constant throughout the motion and is given by:
\[ v_x = u \cos \theta \]

5. **Find the Resultant Velocity**:
The resultant velocity \( v \) at half of the maximum height can be calculated using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values of \( v_x \) and \( v_y \):
\[ v = \sqrt{(u \cos \theta)^2 + \left(\frac{u \sin \theta}{\sqrt{2}}\right)^2} \]
This simplifies to:
\[ v = \sqrt{u^2 \cos^2 \theta + \frac{u^2 \sin^2 \theta}{2}} = u \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}} \]

6. **Final Simplification**:
We can express this as:
\[ v = u \sqrt{\frac{2 \cos^2 \theta + \sin^2 \theta}{2}} = u \sqrt{\frac{1 + \cos 2\theta}{2}} \]

### Final Answer:
Thus, the velocity of the particle at half of its maximum height is:
\[ v = u \sqrt{\frac{1 + \cos 2\theta}{2}} \]