Question

A projectile is fired at an angle of $45^\circ$ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection, is

• A. $ 45^\circ$
• B. $60^\circ$
• C. $ \tan^{ - 1} \frac{1}{2}$
• D. $ \tan^{ - 1} \bigg( \frac{\sqrt 3}{2}\bigg)$

Answer 1

The Correct Option is C

To find the elevation angle of the projectile at its highest point as seen from the point of projection, we need to understand the motion of the projectile. 

1. **Determine the Y-Coordinate at the Highest Point:**
When the projectile is fired at an angle of \(45^\circ\) with an initial velocity \(u\), the vertical and horizontal components of the velocity are equal. At the highest point, the vertical component of the projectile's velocity becomes zero due to gravity, while the horizontal component remains constant.

2. **Calculate the Range and Maximum Height:**
The range \(R\) and maximum height \(H\) of a projectile are given by the formulas:

• Range \(R = \frac{u^2 \sin 2\theta}{g}\)
• Maximum Height \(H = \frac{u^2 \sin^2\theta}{2g}\)

For a launch angle \(45^\circ\):

• Range \(R = \frac{u^2}{g}\)
• Maximum Height \(H = \frac{u^2}{4g}\)

 

3. **Elevation Angle at the Highest Point:**
At the highest point, the coordinates of the projectile are \(\left(\frac{R}{2}, H\right)\) or \(\left(\frac{u^2}{2g}, \frac{u^2}{4g}\right)\). To find the elevation angle of the projectile from the point of projection, we calculate the angle \(\theta\) that the line connecting the origin (0,0) to the highest point makes with the horizontal axis.

The tangent of this angle is given by:

\[ \tan \theta = \frac{\text{Vertical Component}}{\text{Horizontal Component}} = \frac{H}{\frac{R}{2}} = \frac{\frac{u^2}{4g}}{\frac{u^2}{2g}} = \frac{1}{2} \]

Therefore, the elevation angle is \(\theta = \tan^{-1}\left(\frac{1}{2}\right)\). This matches the given correct answer.

 

4. **Conclusion:**
Thus, the correct answer is:

\(\tan^{-1} \frac{1}{2}\)