A shell fired from the ground is just able to cross horizontally the top of a wall `90 m` away and `45 m` high. The direction of projection of the shell will be.

Options

A. `25^@`
B. `30^@`
C. `60^@`
D. `45^@`

Explanation

To find the direction of projection of a shell fired from the ground that just crosses the top of a wall 90 m away and 45 m high, we can use the principles of projectile motion.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Distance to the wall (horizontal range, R) = 90 m
- Height of the wall (maximum height, H) = 45 m

2. **Understanding the Projectile Motion:**
- The shell crosses the wall horizontally at its maximum height. This means that at the point of crossing, the vertical component of the velocity is zero.

3. **Using the Range Formula:**
- The range of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
- Where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

4. **Using the Maximum Height Formula:**
- The maximum height reached by the projectile is given by:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]

5. **Set Up the Equations:**
- From the maximum height equation, we can express \( u^2 \sin^2(\theta) \):
\[ u^2 \sin^2(\theta) = 2gH \]
- Substitute \( H = 45 \, \text{m} \):
\[ u^2 \sin^2(\theta) = 2g \cdot 45 \]

6. **Relate the Two Equations:**
- From the range equation, we can express \( u^2 \sin(2\theta) \):
\[ u^2 \sin(2\theta) = gR \]
- Substitute \( R = 90 \, \text{m} \):
\[ u^2 \sin(2\theta) = g \cdot 90 \]

7. **Using the Identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \):**
- Substitute this into the range equation:
\[ u^2 \cdot 2 \sin(\theta) \cos(\theta) = g \cdot 90 \]

8. **Divide the Two Equations:**
- Divide the maximum height equation by the range equation:
\[ \frac{u^2 \sin^2(\theta)}{u^2 \cdot 2 \sin(\theta) \cos(\theta)} = \frac{2gH}{gR} \]
- This simplifies to:
\[ \frac{\sin(\theta)}{2 \cos(\theta)} = \frac{H}{R} \]
- Substitute \( H = 45 \) and \( R = 90 \):
\[ \frac{\sin(\theta)}{2 \cos(\theta)} = \frac{45}{90} = \frac{1}{2} \]

9. **Solve for \( \tan(\theta) \):**
- Rearranging gives:
\[ \tan(\theta) = \frac{1}{2} \]

10. **Find the Angle \( \theta \):**
- Using the arctangent function:
\[ \theta = \tan^{-1}\left(\frac{1}{2}\right) \]

### Conclusion:
The direction of projection of the shell is given by:
\[ \theta \approx 26.57^\circ \]