If the time of flight of a bullet over a horizontal range R is T, then the angle of projection with horizontal is -

Options

A. `tan^(-1)((gT^(2))/(2R))`
B. `tan^(-1)((2R^(2))/(gT))`
C. `tan^(-1)((2R)/(g^(2)T))`
D. `tan^(-1)((2R)/(gT))`

Answer

To find the angle of projection \(\theta\) with the horizontal when the time of flight \(T\) and the horizontal range \(R\) of a projectile are given, we can use the following steps:

### Step 1: Understand the equations of motion for projectile
The time of flight \(T\) for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
where \(u\) is the initial velocity, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity.

### Step 2: Express initial velocity \(u\) in terms of \(T\) and \(\theta\)
From the time of flight equation, we can rearrange it to find \(u\):
\[ u = \frac{gT}{2\sin \theta} \]

### Step 3: Use the range formula for projectile motion
The horizontal range \(R\) of the projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Substituting the expression for \(u\) from Step 2 into the range formula:
\[ R = \frac{\left(\frac{gT}{2\sin \theta}\right)^2 \sin 2\theta}{g} \]

### Step 4: Simplify the range equation
Substituting \(u\) into the range equation:
\[ R = \frac{g^2 T^2 \sin 2\theta}{4 \sin^2 \theta g} \]
This simplifies to:
\[ R = \frac{gT^2 \sin 2\theta}{4 \sin^2 \theta} \]

### Step 5: Rearranging the equation
Rearranging the equation gives:
\[ \sin 2\theta = \frac{4R \sin^2 \theta}{gT^2} \]

### Step 6: Use the identity for \(\sin 2\theta\)
Using the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), we can substitute:
\[ 2 \sin \theta \cos \theta = \frac{4R \sin^2 \theta}{gT^2} \]

### Step 7: Cancel \(\sin \theta\) and rearrange
Assuming \(\sin \theta \neq 0\), we can cancel one \(\sin \theta\):
\[ 2 \cos \theta = \frac{4R \sin \theta}{gT^2} \]
Rearranging gives:
\[ \tan \theta = \frac{gT^2}{2R} \]

### Step 8: Find the angle of projection
Finally, we can express \(\theta\) as:
\[ \theta = \tan^{-1}\left(\frac{gT^2}{2R}\right) \]

### Conclusion
Thus, the angle of projection \(\theta\) with the horizontal is:
\[ \theta = \tan^{-1}\left(\frac{gT^2}{2R}\right) \]