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

### Step 1: Determine the maximum height of the projectile

The maximum height \( h \) of a projectile is given by the formula:
\[ h = \frac{v^2 \sin^2 \theta}{2g} \]
where \( g \) is the acceleration due to gravity.

### Step 2: Calculate half the maximum height

Half of the maximum height is:
\[ \frac{h}{2} = \frac{1}{2} \cdot \frac{v^2 \sin^2 \theta}{2g} = \frac{v^2 \sin^2 \theta}{4g} \]

### Step 3: Set up the equations of motion

The vertical position \( y \) of the projectile as a function of time \( t \) is given by:
\[ y = v \sin \theta \cdot t - \frac{1}{2} g t^2 \]

### Step 4: Find the time when the projectile reaches half the maximum height

Set \( y = \frac{h}{2} \):
\[ \frac{v^2 \sin^2 \theta}{4g} = v \sin \theta \cdot t - \frac{1}{2} g t^2 \]
Rearranging gives us a quadratic equation in \( t \):
\[ \frac{1}{2} g t^2 - v \sin \theta \cdot t + \frac{v^2 \sin^2 \theta}{4g} = 0 \]

### Step 5: Solve the quadratic equation for time

Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = \frac{1}{2} g \), \( b = -v \sin \theta \), and \( c = \frac{v^2 \sin^2 \theta}{4g} \).

Calculating the discriminant:
\[ b^2 - 4ac = (v \sin \theta)^2 - 4 \cdot \frac{1}{2} g \cdot \frac{v^2 \sin^2 \theta}{4g} = v^2 \sin^2 \theta - \frac{v^2 \sin^2 \theta}{2} = \frac{v^2 \sin^2 \theta}{2} \]

Thus, the times when the projectile reaches half the maximum height are:
\[ t_1, t_2 = \frac{v \sin \theta \pm \sqrt{\frac{v^2 \sin^2 \theta}{2}}}{g} \]

### Step 6: Calculate the average velocity

The average velocity \( v_{avg} \) between the two times \( t_1 \) and \( t_2 \) is given by:
\[ v_{avg} = \frac{x_2 - x_1}{t_2 - t_1} \]
Where \( x_1 \) and \( x_2 \) are the horizontal displacements at times \( t_1 \) and \( t_2 \).

### Step 7: Find the horizontal displacement

The horizontal displacement \( x \) is given by:
\[ x = v \cos \theta \cdot t \]
Thus,
\[ x_1 = v \cos \theta \cdot t_1, \quad x_2 = v \cos \theta \cdot t_2 \]

### Step 8: Substitute and simplify

Substituting into the average velocity formula:
\[ v_{avg} = \frac{v \cos \theta (t_2 - t_1)}{t_2 - t_1} \]
This simplifies to:
\[ v_{avg} = v \cos \theta \]

### Conclusion

The average velocity of the projectile between the instants it crosses half the maximum height is:
\[ v_{avg} = v \sin \theta \]
This average velocity is horizontal and in the plane of projection.