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. `30^@`
B. `45^@`
C. `60^@`
D. `70^@`

Explanation

To solve the problem, we need to find the angle at which a shot is fired from a point 200 meters away from the base of a tower that is 100 meters high, such that the shot just passes over the tower horizontally.

### 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 \( R = 200 \, \text{m} \).
- We need to find the angle \( \theta \) at which the shot is fired.

2. **Projectile Motion Basics:**
- The shot will follow a projectile motion path.
- The horizontal distance covered by the projectile when it reaches the height of the tower is equal to the range of the projectile.

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

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

5. **Setting Up the Equations:**
- We have two equations:
1. \( R = \frac{U^2 \sin(2\theta)}{g} \)
2. \( H = \frac{U^2 \sin^2(\theta)}{2g} \)

6. **Expressing \( U^2 \sin^2(\theta) \):**
- From the height equation, we can express \( U^2 \sin^2(\theta) \):
\[ U^2 \sin^2(\theta) = 2gH \]

7. **Substituting into the Range Equation:**
- Substitute \( U^2 \sin^2(\theta) \) into the range equation:
\[ R = \frac{2gH \sin(2\theta)}{g \sin^2(\theta)} \]
- Simplifying gives:
\[ R = \frac{2H \sin(2\theta)}{\sin^2(\theta)} \]

8. **Using the Identity for \( \sin(2\theta) \):**
- Recall that \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \):
\[ R = \frac{4H \sin(\theta) \cos(\theta)}{\sin^2(\theta)} \]
- This simplifies to:
\[ R = 4H \cot(\theta) \]

9. **Finding \( \tan(\theta) \):**
- Rearranging gives:
\[ \cot(\theta) = \frac{R}{4H} \]
- Therefore:
\[ \tan(\theta) = \frac{4H}{R} \]

10. **Substituting Values:**
- Substitute \( H = 100 \, \text{m} \) and \( R = 200 \, \text{m} \):
\[ \tan(\theta) = \frac{4 \times 100}{200} = 2 \]

11. **Finding \( \theta \):**
- Now, calculate \( \theta \):
\[ \theta = \tan^{-1}(2) \]

12. **Final Answer:**
- The angle \( \theta \) at which the shot is fired is approximately \( 63.4^\circ \).