The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to below in the direction of motion of the projectile, giving it a constant horizontal acceleration `=g//2`. Under the same conditions of projection. Find the horizontal range of the projectile.

Options

A. `R+H`
B. `R+2H`
C. `R`
D. `R+H//2`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the initial conditions
We know that the horizontal range of a projectile is \( R \) and the maximum height attained is \( H \). The initial velocity \( u \) can be resolved into horizontal and vertical components:
- Horizontal component: \( u \cos \theta \)
- Vertical component: \( u \sin \theta \)

### Step 2: Calculate the time of flight
The time of flight \( T \) for a projectile is given by:
\[ T = \frac{2u \sin \theta}{g} \]

### Step 3: Express the range in terms of initial velocity and angle
The range \( R \) can be expressed as:
\[ R = u \cos \theta \cdot T = u \cos \theta \cdot \frac{2u \sin \theta}{g} \]
This simplifies to:
\[ R = \frac{2u^2 \sin \theta \cos \theta}{g} \]

### Step 4: Relate the maximum height to initial velocity and angle
The maximum height \( H \) is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

### Step 5: Introduce the effect of wind
Now, a wind blows in the direction of motion, providing a constant horizontal acceleration of \( \frac{g}{2} \). We need to find the new range \( R' \).

### Step 6: Use the equations of motion
The new horizontal distance covered due to the wind can be calculated using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( u = u \cos \theta \), \( a = \frac{g}{2} \), and \( t = T = \frac{2u \sin \theta}{g} \).

### Step 7: Substitute values into the equation
Substituting the values into the equation:
\[ R' = u \cos \theta \cdot T + \frac{1}{2} \cdot \frac{g}{2} \cdot T^2 \]
Substituting \( T \):
\[ R' = u \cos \theta \cdot \frac{2u \sin \theta}{g} + \frac{1}{2} \cdot \frac{g}{2} \cdot \left(\frac{2u \sin \theta}{g}\right)^2 \]

### Step 8: Simplify the equation
Calculating the second term:
\[ \frac{1}{2} \cdot \frac{g}{2} \cdot \left(\frac{2u \sin \theta}{g}\right)^2 = \frac{g}{4} \cdot \frac{4u^2 \sin^2 \theta}{g^2} = \frac{u^2 \sin^2 \theta}{g} \]
Now, substituting back:
\[ R' = \frac{2u^2 \sin \theta \cos \theta}{g} + \frac{u^2 \sin^2 \theta}{g} \]

### Step 9: Factor the equation
Factoring out \( \frac{u^2}{g} \):
\[ R' = \frac{u^2}{g} (2 \sin \theta \cos \theta + \sin^2 \theta) \]
Using the identity \( 2 \sin \theta \cos \theta = \sin(2\theta) \):
\[ R' = \frac{u^2}{g} \left(\sin(2\theta) + \sin^2 \theta\right) \]

### Step 10: Relate back to original range and height
From the earlier steps, we know:
- \( R = \frac{2u^2 \sin \theta \cos \theta}{g} \)
- \( H = \frac{u^2 \sin^2 \theta}{2g} \)

Thus, the new range can be expressed as:
\[ R' = R + H \]

### Conclusion
The new horizontal range of the projectile when the wind is blowing is:
\[ R' = R + H \]