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 velocity of the projectile when it is at a height equal to half of the maximum height is.

Options

A. `v sqrt(cos^2 theta + (sin^2 theta)/(2))`
B. `sqrt(2) v cos theta`
C. `sqrt(2) v sin theta`
D. `v tan theta sec theta`

Explanation

To solve the problem, we need to find the velocity of a projectile when it is at a height equal to half of its maximum height. Let's break it down step by step.

### Step 1: Determine the maximum height
The maximum height \( H \) reached by a projectile thrown with an initial velocity \( v \) at an angle \( \theta \) 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 of the maximum height
Half of the maximum height \( h \) is:
\[ h = \frac{H}{2} = \frac{1}{2} \cdot \frac{v^2 \sin^2 \theta}{2g} = \frac{v^2 \sin^2 \theta}{4g} \]

### Step 3: Use the vertical motion equation
We will use the kinematic equation for vertical motion to find the vertical component of velocity \( V_y \) when the projectile is at height \( h \):
\[ V_y^2 = V_{y0}^2 - 2g h \]
where \( V_{y0} = v \sin \theta \) is the initial vertical velocity component.

Substituting \( h \) into the equation:
\[ V_y^2 = (v \sin \theta)^2 - 2g \left(\frac{v^2 \sin^2 \theta}{4g}\right) \]

### Step 4: Simplify the equation
Now, substituting and simplifying:
\[ V_y^2 = v^2 \sin^2 \theta - \frac{v^2 \sin^2 \theta}{2} = \frac{v^2 \sin^2 \theta}{2} \]

### Step 5: Calculate the horizontal component of velocity
The horizontal component of velocity \( V_x \) remains constant throughout the projectile's motion:
\[ V_x = v \cos \theta \]

### Step 6: Find the resultant velocity
The resultant velocity \( V \) when the projectile is at height \( h \) can be found using the Pythagorean theorem:
\[ V = \sqrt{V_x^2 + V_y^2} \]
Substituting the values:
\[ V = \sqrt{(v \cos \theta)^2 + \left(\sqrt{\frac{v^2 \sin^2 \theta}{2}}\right)^2} \]
\[ V = \sqrt{(v \cos \theta)^2 + \frac{v^2 \sin^2 \theta}{2}} \]

### Step 7: Factor out \( v^2 \)
Factoring out \( v^2 \):
\[ V = v \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}} \]

### Step 8: Final expression
Thus, the final expression for the velocity of the projectile at half of its maximum height is:
\[ V = v \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{2}} \]