A projectile is fired at an angle of `45^(@)` with the horizontal. Elevation angle of the projection at its highest point as seen from the point of projection is

Options

A. `60^(@)`
B. `tan^(-1)(1/2)`
C. `tan^(-1)(sqrt(3)/2)`
D. `45^(@)`

Explanation

To solve the problem of finding the elevation angle of the projectile at its highest point as seen from the point of projection, we can follow these steps:

### Step 1: Understand the Projectile Motion
When a projectile is fired at an angle of 45 degrees, it follows a parabolic trajectory. The highest point of the projectile's path is known as the apex.

### Step 2: Determine the Components of Velocity
At the time of projection, the initial velocity can be resolved into two components:
- Horizontal component (Vx) = V * cos(45°)
- Vertical component (Vy) = V * sin(45°)

Since sin(45°) = cos(45°) = 1/√2, we have:
- Vx = V/√2
- Vy = V/√2

### Step 3: Identify the Highest Point
At the highest point of the projectile's motion, the vertical component of the velocity becomes zero (Vy = 0). However, the horizontal component (Vx) remains constant throughout the motion.

### Step 4: Analyze the Position at the Highest Point
At the highest point, the projectile is at its maximum height. From the point of projection, we need to find the angle of elevation to this highest point.

### Step 5: Use Geometry to Find the Angle
Let’s denote:
- H = maximum height reached by the projectile
- R = horizontal range of the projectile

At the highest point, the horizontal distance from the point of projection is R/2 (since the range is symmetric and the highest point is at the midpoint of the range).

Using the tangent function in the right triangle formed by the height and horizontal distance:
\[ \tan(\theta) = \frac{H}{R/2} \]

### Step 6: Find the Relationship Between H and R
For a projectile launched at angle θ, the maximum height (H) can be given by:
\[ H = \frac{V^2 \sin^2(θ)}{2g} \]
And the range (R) is given by:
\[ R = \frac{V^2 \sin(2θ)}{g} \]
Substituting θ = 45° (where sin(2*45°) = 1):
\[ R = \frac{V^2}{g} \]
Thus, at θ = 45°, we can express H:
\[ H = \frac{V^2 \cdot (1/2)}{2g} = \frac{V^2}{4g} \]

### Step 7: Substitute H and R into the Tangent Equation
Now substituting H and R into the tangent equation:
\[ \tan(\theta) = \frac{H}{R/2} = \frac{\frac{V^2}{4g}}{\frac{V^2}{2g}} = \frac{1}{2} \]

### Step 8: Calculate the Angle
To find the angle θ:
\[ \theta = \tan^{-1}(1/2) \]

### Conclusion
The elevation angle of the projectile at its highest point as seen from the point of projection is:
\[ \theta = \tan^{-1}(1/2) \]