A projectile is fired from level ground at an angle `theta` above the horizontal. The elevation angle `phi` of the highest point as seen from the launch point is related to `theta` by the relation.

Options

A. `tan phi = 2 tan theta`
B. `tan phi = tan theta`
C. `tan phi = (1)/(2) tan theta`
D. `tan phi = (1)/(4) tan theta`

Explanation

To find the relation between the launch angle \( \theta \) and the elevation angle \( \phi \) of the highest point of a projectile as seen from the launch point, we can follow these steps:

### Step 1: Understand the projectile motion
When a projectile is launched from the ground at an angle \( \theta \), it follows a parabolic trajectory. The highest point of the projectile is called the apex.

### Step 2: Determine the coordinates of the highest point
The horizontal and vertical components of the initial velocity \( v_0 \) can be expressed as:
- \( v_{0x} = v_0 \cos(\theta) \)
- \( v_{0y} = v_0 \sin(\theta) \)

At the highest point, the vertical component of the velocity becomes zero. The time \( t \) to reach the highest point can be calculated using the formula:
\[ t = \frac{v_{0y}}{g} = \frac{v_0 \sin(\theta)}{g} \]
where \( g \) is the acceleration due to gravity.

### Step 3: Calculate the height at the highest point
The height \( H \) of the projectile at the highest point can be calculated using the formula:
\[ H = v_{0y} t - \frac{1}{2} g t^2 \]
Substituting \( t \) from the previous step:
\[ H = v_0 \sin(\theta) \left(\frac{v_0 \sin(\theta)}{g}\right) - \frac{1}{2} g \left(\frac{v_0 \sin(\theta)}{g}\right)^2 \]
This simplifies to:
\[ H = \frac{v_0^2 \sin^2(\theta)}{2g} \]

### Step 4: Determine the horizontal distance at the highest point
The horizontal distance \( R \) covered by the projectile when it reaches the highest point can be calculated as:
\[ R = v_{0x} \cdot t = v_0 \cos(\theta) \cdot \frac{v_0 \sin(\theta)}{g} = \frac{v_0^2 \sin(\theta) \cos(\theta)}{g} \]

### Step 5: Relate the angles \( \theta \) and \( \phi \)
The angle \( \phi \) is the angle of elevation from the launch point to the highest point. Using the tangent function, we can express \( \tan(\phi) \) as:
\[ \tan(\phi) = \frac{H}{R} \]
Substituting the expressions for \( H \) and \( R \):
\[ \tan(\phi) = \frac{\frac{v_0^2 \sin^2(\theta)}{2g}}{\frac{v_0^2 \sin(\theta) \cos(\theta)}{g}} = \frac{\sin(\theta)}{2 \cos(\theta)} \]
Thus, we have:
\[ \tan(\phi) = \frac{1}{2} \tan(\theta) \]

### Conclusion
The relation between the elevation angle \( \phi \) of the highest point and the launch angle \( \theta \) is given by:
\[ \tan(\phi) = \frac{1}{2} \tan(\theta) \]