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. `45^(@)`
B. `60^(@)`
C. `tan^(-1)(1)/(2)`
D. `tan^(-1)((sqrt3)/(2))`

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-by-Step Solution:

1. **Understanding the Problem**:
A projectile is fired at an angle of \( 45^\circ \) with the horizontal. We need to find the elevation angle \( \alpha \) at the highest point of the projectile's trajectory as viewed from the point of projection.

2. **Identify Key Points**:
- Let \( O \) be the point of projection.
- Let \( A \) be the highest point of the projectile's trajectory.
- The horizontal distance from \( O \) to the point directly below \( A \) is \( \frac{R}{2} \) (where \( R \) is the range of the projectile).

3. **Determine Maximum Height**:
The maximum height \( h_{max} \) of a projectile is given by the formula:
\[ h_{max} = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

4. **Relate Height and Range**:
The range \( R \) of the projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
For \( \theta = 45^\circ \), \( \sin(90^\circ) = 1 \), thus:
\[ R = \frac{u^2}{g} \]

5. **Substituting Values**:
From the above formulas, we can express \( h_{max} \) in terms of \( R \):
\[ h_{max} = \frac{R}{4} \]

6. **Using Right Triangle**:
To find the elevation angle \( \alpha \), we can use the right triangle formed by the height \( h_{max} \) and the horizontal distance \( \frac{R}{2} \):
\[ \tan \alpha = \frac{h_{max}}{\frac{R}{2}} = \frac{\frac{R}{4}}{\frac{R}{2}} = \frac{1}{2} \]

7. **Finding the Angle**:
Now, we can find \( \alpha \):
\[ \alpha = \tan^{-1}\left(\frac{1}{2}\right) \]

### Final Answer:
Thus, the elevation angle \( \alpha \) at the highest point as seen from the point of projection is:
\[ \alpha = \tan^{-1}\left(\frac{1}{2}\right) \]