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))/(Rg))`
B. `cos^(-1)((Rg)/(v^(2)))`
C. `(1)/(2)sin^(-1)((v^(2))/(Rg))`
D. `(1)/(2)sin^(-1)((Rg)/(v^(2)))`

Explanation

To find the angle of elevation of a gun that fires a shell with a muzzle speed \( V \) and has a range \( R \), we can use the formula for the range of a projectile. The formula for the range \( R \) of a projectile launched at an angle \( \theta \) with an initial velocity \( V \) is given by:

\[ R = \frac{V^2 \sin 2\theta}{g} \]

where:
- \( R \) is the range,
- \( V \) is the muzzle speed,
- \( g \) is the acceleration due to gravity,
- \( \theta \) is the angle of elevation.

### Step 1: Rearranging the Range Formula
We can rearrange the formula to solve for \( \sin 2\theta \):

\[ \sin 2\theta = \frac{R \cdot g}{V^2} \]

### Step 2: Finding the Angle \( \theta \)
To find \( \theta \), we take the inverse sine (arcsin) of both sides:

\[ 2\theta = \sin^{-1}\left(\frac{R \cdot g}{V^2}\right) \]

### Step 3: Solving for \( \theta \)
Now, we divide both sides by 2 to isolate \( \theta \):

\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{R \cdot g}{V^2}\right) \]

### Conclusion
Thus, the angle of elevation \( \theta \) of the gun is given by:

\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{R \cdot g}{V^2}\right) \]