The ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is x. The angle of projection with the horizontal is

Options

A. `sin^(1) x`
B. `cos^(-1)x`
C. `sin^(-1) (1//x)`
D. `cos^(-1) (1//x)`

Explanation

To solve the problem, we need to find the angle of projection (θ) with the horizontal given the ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is x.

### Step-by-Step Solution:

1. **Define Variables**:
- Let the speed of the projectile at the point of projection be \( u \).
- The speed at the top of the trajectory is given by \( u \cos \theta \), where \( \theta \) is the angle of projection.

2. **Set Up the Ratio**:
- According to the problem, the ratio of the speed at the point of projection to the speed at the top of the trajectory is \( x \).
- This can be expressed mathematically as:
\[ \frac{u}{u \cos \theta} = x \]

3. **Simplify the Equation**:
- The \( u \) terms cancel out:
\[ \frac{1}{\cos \theta} = x \]

4. **Rearranging the Equation**:
- From the above equation, we can express \( \cos \theta \):
\[ \cos \theta = \frac{1}{x} \]

5. **Finding the Angle of Projection**:
- To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{1}{x}\right) \]

6. **Conclusion**:
- Thus, the angle of projection with the horizontal is:
\[ \theta = \cos^{-1}\left(\frac{1}{x}\right) \]

### Final Answer:
The angle of projection with the horizontal is \( \cos^{-1}\left(\frac{1}{x}\right) \).

---