A projectile is thrown with velocity `v` at an angle `theta` with the horizontal. When the projectile is at a height equal to half of the maximum height,.
The vertical component of the velocity of projectile is.

Options

A. `3 v sin theta`
B. `v sin theta`
C. `(v sin theta)/(sqrt(2))`
D. `(v sin theta)/(sqrt(3))`

Explanation

To find the vertical component of the velocity of a projectile at a height equal to half of its maximum height, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- A projectile is thrown with an initial velocity \( v \) at an angle \( \theta \) with the horizontal.
- The vertical component of the initial velocity \( v_y \) is given by:
\[ v_y = v \sin \theta \]

2. **Determine the Maximum Height**:
- The maximum height \( H \) reached by the projectile can be calculated using the formula:
\[ H = \frac{v^2 \sin^2 \theta}{2g} \]
where \( g \) is the acceleration due to gravity.

3. **Calculate Half of the Maximum Height**:
- Half of the maximum height \( h \) is:
\[ h = \frac{H}{2} = \frac{1}{2} \cdot \frac{v^2 \sin^2 \theta}{2g} = \frac{v^2 \sin^2 \theta}{4g} \]

4. **Apply the Equation of Motion**:
- We can use the third equation of motion for the vertical direction:
\[ v_y^2 = v_{y0}^2 - 2gh \]
where \( v_{y0} = v \sin \theta \) is the initial vertical velocity.

5. **Substituting Values**:
- Substitute \( v_{y0} \) and \( h \) into the equation:
\[ v_y^2 = (v \sin \theta)^2 - 2g \left(\frac{v^2 \sin^2 \theta}{4g}\right) \]

6. **Simplifying the Equation**:
- Simplifying gives:
\[ v_y^2 = v^2 \sin^2 \theta - \frac{v^2 \sin^2 \theta}{2} \]
- This simplifies to:
\[ v_y^2 = \frac{v^2 \sin^2 \theta}{2} \]

7. **Finding the Vertical Component of Velocity**:
- Taking the square root of both sides gives:
\[ v_y = \frac{v \sin \theta}{\sqrt{2}} \]

### Final Answer:
The vertical component of the velocity of the projectile when it is at a height equal to half of the maximum height is:
\[ v_y = \frac{v \sin \theta}{\sqrt{2}} \]