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 derive the relation between the elevation angle \( \phi \) of the highest point of a projectile as seen from the launch point and the angle of projection \( \theta \), we will 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 maximum height \( H \) and the horizontal range \( R \) are key components of this motion.

### Step 2: Determine the maximum height \( H \)
The maximum height \( H \) of the projectile can be calculated using the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial velocity and \( g \) is the acceleration due to gravity.

### Step 3: Determine the horizontal range \( R \)
The total horizontal range \( R \) of the projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]

### Step 4: Find the horizontal distance at maximum height
At the maximum height, the horizontal distance traveled by the projectile is half of the total range:
\[ \text{Horizontal distance at max height} = \frac{R}{2} = \frac{u^2 \sin 2\theta}{2g} \]

### Step 5: Relate elevation angle \( \phi \) to \( H \) and horizontal distance
The elevation angle \( \phi \) at the maximum height can be expressed using the tangent function:
\[ \tan \phi = \frac{H}{\text{Horizontal distance at max height}} = \frac{H}{\frac{R}{2}} \]

### Step 6: Substitute the values of \( H \) and \( R \)
Substituting the expressions for \( H \) and \( R \):
\[ \tan \phi = \frac{\frac{u^2 \sin^2 \theta}{2g}}{\frac{u^2 \sin 2\theta}{2g}} = \frac{\sin^2 \theta}{\sin 2\theta} \]

### Step 7: Simplify the expression
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ \tan \phi = \frac{\sin^2 \theta}{2 \sin \theta \cos \theta} = \frac{\sin \theta}{2 \cos \theta} = \frac{1}{2} \tan \theta \]

### Step 8: Final relation
Thus, we can express \( \tan \phi \) in terms of \( \tan \theta \):
\[ \tan \phi = \tan \frac{\theta}{2} \]

### Conclusion
The relation between the elevation angle \( \phi \) at the maximum height and the angle of projection \( \theta \) is:
\[ \tan \phi = \tan \frac{\theta}{2} \]