If the range of a gun which fires a shell with muzzle speed V is R , then the angle of elevation of the gun is

Options

A. ` cos ^(-1) (V^2/(gR))`
B. `cos ^(-1) ((gR)/(v^2))`
C. `1/2 sin ^(-1) ((v^2)/(gR))`
D. `1/2 sin ^(-1) ((gR)/(v^2))`

Explanation

To find the angle of elevation (θ) of a gun that fires a shell with a muzzle speed (V) and achieves a range (R), we can use the formula for the range of projectile motion.

### Step-by-Step Solution:

1. **Understand the Range Formula**:
The range (R) of a projectile launched with an initial velocity (V) at an angle (θ) is given by the formula:
\[ R = \frac{V^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity.

2. **Rearranging the Formula**:
We need to express \( \sin(2\theta) \) in terms of R, V, and g. Rearranging the range formula gives:
\[ \sin(2\theta) = \frac{gR}{V^2} \]

3. **Finding 2θ**:
To find \( 2\theta \), we take the inverse sine (arcsin) of both sides:
\[ 2\theta = \sin^{-1}\left(\frac{gR}{V^2}\right) \]

4. **Calculating θ**:
Finally, we divide by 2 to find θ:
\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{gR}{V^2}\right) \]

5. **Conclusion**:
Thus, the angle of elevation of the gun is:
\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{gR}{V^2}\right) \]

### Final Answer:
The angle of elevation of the gun is:
\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{gR}{V^2}\right) \]