A projectile is thrown with a speed v at an angle `theta` with the vertical. Its average velocity between the instants it crosses half the maximum height is

Options

A. `v sin theta`, horizontal and in the plane of projection
B. `v cos theta, `horizontal and in the plane of projection
C. `2 v sin theta, `horizontal and perpendicular to the plane of projection
D. `2 v cos theta`, vertical and in the plane of projection

Explanation

To solve the problem of finding the average velocity of a projectile thrown with speed \( v \) at an angle \( \theta \) with the vertical between the instants it crosses half the maximum height, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Projectile Motion**:
- The projectile is thrown at an angle \( \theta \) with respect to the vertical. This means it makes an angle \( 90^\circ - \theta \) with the horizontal.
- The motion can be analyzed in terms of horizontal and vertical components.

2. **Components of Velocity**:
- The initial velocity \( v \) can be broken down into its components:
- Horizontal component: \( v_x = v \sin(90^\circ - \theta) = v \cos(\theta) \)
- Vertical component: \( v_y = v \cos(90^\circ - \theta) = v \sin(\theta) \)

3. **Maximum Height Calculation**:
- The maximum height \( H \) reached by the projectile can be calculated using the formula:
\[ H = \frac{v_y^2}{2g} = \frac{(v \sin(\theta))^2}{2g} \]
- Half of the maximum height is \( \frac{H}{2} = \frac{(v \sin(\theta))^2}{4g} \).

4. **Time to Reach Half Maximum Height**:
- The time \( t \) to reach half the maximum height can be found using the equation of motion:
\[ h = v_y t - \frac{1}{2} g t^2 \]
Setting \( h = \frac{H}{2} \) and solving for \( t \):
\[ \frac{(v \sin(\theta))^2}{4g} = (v \sin(\theta)) t - \frac{1}{2} g t^2 \]
- This quadratic equation can be solved for \( t \).

5. **Finding Average Velocity**:
- The average velocity \( \bar{v} \) between the two instances when the projectile crosses half the maximum height can be calculated as:
\[ \bar{v} = \frac{v_1 + v_2}{2} \]
- Since the horizontal component of velocity remains constant, we have:
\[ v_1 = v_x \quad \text{and} \quad v_2 = v_x \]
- The vertical components at these two points will be equal in magnitude but opposite in direction, so they will cancel out when calculating the average.

6. **Final Expression for Average Velocity**:
- Since the horizontal components are the same and the vertical components cancel out, the average velocity simplifies to:
\[ \bar{v} = v_x = v \cos(\theta) \]

### Final Result:
The average velocity of the projectile between the instants it crosses half the maximum height is given by:
\[ \bar{v} = v \cos(\theta) \]