An arrow shots from a bow with velocity v at an angle ` theta` with the horizontal range R . Its range when it is projected at angle ` 2 theta` with the same velocity is

Options

A. 2R
B. `R/2`
C. `2R cos 2theta`
D. `R/(2cos^(2) theta)`

Answer

To solve the problem of finding the range of an arrow shot at an angle of \(2\theta\) with the same initial velocity \(v\), we can follow these steps:

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

### Step 2: Write the expression for the initial range
Given that the range when the arrow is shot at angle \(\theta\) is \(R\), we can express this as:
\[ R = \frac{v^2 \sin(2\theta)}{g} \]

### Step 3: Determine the new range when the angle is \(2\theta\)
Now, we want to find the new range \(R'\) when the angle of projection is changed to \(2\theta\). Using the range formula again, we have:
\[ R' = \frac{v^2 \sin(2 \cdot 2\theta)}{g} = \frac{v^2 \sin(4\theta)}{g} \]

### Step 4: Use the double angle identity for sine
Using the double angle identity for sine, we can express \(\sin(4\theta)\) as:
\[ \sin(4\theta) = 2 \sin(2\theta) \cos(2\theta) \]
Substituting this into the expression for \(R'\):
\[ R' = \frac{v^2 \cdot 2 \sin(2\theta) \cos(2\theta)}{g} \]

### Step 5: Relate \(R'\) to \(R\)
Now, we can relate \(R'\) to \(R\):
\[ R' = 2 \cdot \frac{v^2 \sin(2\theta)}{g} \cdot \cos(2\theta) = 2R \cos(2\theta) \]

### Conclusion
Thus, the range when the arrow is projected at an angle \(2\theta\) with the same velocity \(v\) is:
\[ R' = 2R \cos(2\theta) \]

### Final Answer
The correct option is \(2R \cos(2\theta)\).

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