A shot is fired from a point at a distance of `200 m` from the foot of a tower `100 m` high so that it just passes over it horizontally. The direction of shot with horizontal is.

Options

A. `75^@`
B. `60^@`
C. `30^@`
D. `45^@`

Explanation

To solve the problem of finding the angle at which a shot is fired horizontally over a tower, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a tower of height \( h = 100 \, \text{m} \).
- The distance from the point of firing to the foot of the tower is \( d = 200 \, \text{m} \).
- The shot is fired horizontally, meaning the initial vertical velocity component is zero.

2. **Using the Projectile Motion Equations**:
- The maximum height \( H \) reached by the projectile can be calculated using the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
- The range \( R \) of the projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]

3. **Setting Up the Equations**:
- Since the shot just passes over the tower, we can set the maximum height \( H \) equal to the height of the tower:
\[ H = 100 \, \text{m} \]
- The horizontal distance to the tower is \( d = 200 \, \text{m} \), which is half the total range when it reaches maximum height.

4. **Relating the Two Equations**:
- From the maximum height equation:
\[ 100 = \frac{u^2 \sin^2 \theta}{2g} \quad \text{(1)} \]
- From the range equation:
\[ 200 = \frac{u^2 \sin 2\theta}{g} \quad \text{(2)} \]

5. **Solving the Equations**:
- From equation (1), we can express \( u^2 \):
\[ u^2 = 200g \sin^2 \theta \quad \text{(3)} \]
- Substitute equation (3) into equation (2):
\[ 200 = \frac{200g \sin^2 \theta \cdot \sin 2\theta}{g} \]
- Simplifying gives:
\[ 1 = \sin^2 \theta \cdot \sin 2\theta \]
- Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ 1 = \sin^2 \theta \cdot 2 \sin \theta \cos \theta \]
- This simplifies to:
\[ 1 = 2 \sin^3 \theta \cos \theta \]

6. **Finding the Angle**:
- Rearranging gives:
\[ \sin^3 \theta \cos \theta = \frac{1}{2} \]
- By trial or using trigonometric identities, we find that \( \theta = 45^\circ \) satisfies this equation since:
\[ \sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2} \]
- Therefore:
\[ \sin^3 45^\circ \cos 45^\circ = \left(\frac{\sqrt{2}}{2}\right)^3 \cdot \frac{\sqrt{2}}{2} = \frac{2\sqrt{2}}{16} = \frac{1}{2} \]

### Final Answer:
The angle of the shot with respect to the horizontal is \( \theta = 45^\circ \).