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

Options

A. `25^(@)`
B. `30^(@)`
C. `45^(@)`
D. none of these

Explanation

To solve the problem of finding the direction of projection of a shell that just crosses the top of a wall, we can follow these steps:

### Step 1: Understand the problem
We have a wall that is 90 meters away from the point of projection and 45 meters high. We need to find the angle of projection (θ) of the shell.

### Step 2: Use the maximum height formula
The maximum height (H) reached by a projectile is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where:
- \( H = 45 \) m (height of the wall)
- \( u \) = initial velocity of the shell
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)

### Step 3: Use the range formula
The range (R) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
In this case, the horizontal distance to the wall is 90 m, so:
\[ R = 90 \, \text{m} \]

### Step 4: Relate the maximum height and range
Since the wall is at half the range (45 m), we can express the maximum height in terms of the range:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
Substituting \( R = 90 \):
\[ R = 2H \]
This means:
\[ 90 = \frac{u^2 \sin 2\theta}{g} \]

### Step 5: Set up the equations
From the maximum height equation:
\[ 45 = \frac{u^2 \sin^2 \theta}{2g} \]
From the range equation:
\[ 90 = \frac{u^2 \sin 2\theta}{g} \]

### Step 6: Solve for \( u^2 \)
From the maximum height equation:
\[ u^2 = \frac{90g}{\sin 2\theta} \]
From the height equation:
\[ u^2 = \frac{90g}{\sin 2\theta} \]

### Step 7: Equate and simplify
Setting the two expressions for \( u^2 \) equal:
\[ \frac{90g}{\sin 2\theta} = 90g \cdot \frac{\sin^2 \theta}{2g} \]
Canceling \( 90g \) from both sides:
\[ 1 = \frac{\sin^2 \theta}{2} \]
Thus:
\[ \sin^2 \theta = 2 \]
This is incorrect, so we need to check our calculations.

### Step 8: Use the relationship between sin and tan
Using the relationship:
\[ \tan \theta = \frac{H}{d} = \frac{45}{90} = \frac{1}{2} \]
Thus:
\[ \theta = \tan^{-1}\left(\frac{1}{2}\right) \]

### Step 9: Calculate the angle
Using a calculator:
\[ \theta \approx 26.57^\circ \]

### Conclusion
The direction of projection of the shell will be approximately \( 26.57^\circ \) above the horizontal.

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