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 when it is at a height equal to half of its maximum height, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The projectile is thrown with an initial velocity \( v \) at an angle \( \theta \) with the horizontal.
- The vertical component of the initial velocity is given by:
\[ V_{y0} = v \sin \theta \]

2. **Determine the Maximum Height**:
- The maximum height \( H \) reached by the projectile can be calculated using the formula:
\[ H = \frac{V_{y0}^2}{2g} = \frac{(v \sin \theta)^2}{2g} \]
- Therefore, half of the maximum height is:
\[ h = \frac{H}{2} = \frac{1}{2} \cdot \frac{(v \sin \theta)^2}{2g} = \frac{(v \sin \theta)^2}{4g} \]

3. **Use the Kinematic Equation**:
- We can use the kinematic equation for vertical motion to find the vertical component of the velocity \( V_y \) when the projectile is at height \( h \):
\[ V_y^2 = V_{y0}^2 - 2gh \]
- Substituting \( V_{y0} \) and \( h \):
\[ V_y^2 = (v \sin \theta)^2 - 2g \left(\frac{(v \sin \theta)^2}{4g}\right) \]

4. **Simplify the Equation**:
- Simplifying the equation:
\[ V_y^2 = (v \sin \theta)^2 - \frac{(v \sin \theta)^2}{2} \]
\[ V_y^2 = \frac{(v \sin \theta)^2}{2} \]

5. **Calculate the Vertical Component of Velocity**:
- Taking the square root to find \( V_y \):
\[ V_y = \sqrt{\frac{(v \sin \theta)^2}{2}} = \frac{v \sin \theta}{\sqrt{2}} \]

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