The horizontal range and maximum height attained by a projectile are R and H respectively. If a constant horizontal acceleration `a=(g)/(4)` is is imparted to the projectile due to wind, then its horizontal range and maximum height will be

Options

A. `(R+H),(H)/(2)`
B. `(R+(H)/(2)),2H`
C. `(R+2H),H`
D. `(R+H),H`

Explanation

To solve the problem, we need to analyze the projectile motion with the given conditions. Let's break it down step by step.

### Step 1: Understanding the Original Conditions
The projectile has an initial horizontal range \( R \) and a maximum height \( H \). The horizontal motion is typically characterized by zero horizontal acceleration, while the vertical motion is influenced by gravitational acceleration \( g \).

### Step 2: Analyzing the New Conditions
Now, we introduce a constant horizontal acceleration \( a = \frac{g}{4} \) due to wind. This changes the dynamics of the horizontal motion but does not affect the vertical motion.

### Step 3: Time of Flight
The time of flight \( T \) for a projectile is given by:
\[ T = \frac{2u \sin \theta}{g} \]
where \( u \) is the initial velocity and \( \theta \) is the angle of projection. This time remains unchanged because the vertical motion is still influenced only by gravity.

### Step 4: Maximum Height
The maximum height \( H \) is given by:
\[ H = \frac{(u \sin \theta)^2}{2g} \]
Since the vertical motion is unaffected by the horizontal acceleration, the maximum height remains \( H \).

### Step 5: New Horizontal Range
The new horizontal range \( R' \) can be calculated using the equations of motion. The formula for horizontal displacement with constant acceleration is:
\[ R' = u_x T + \frac{1}{2} a T^2 \]
where \( u_x = u \cos \theta \) is the horizontal component of the initial velocity.

Substituting the values:
\[ R' = (u \cos \theta) T + \frac{1}{2} \left(\frac{g}{4}\right) T^2 \]
Now, substituting \( T = \frac{2u \sin \theta}{g} \):
\[ R' = (u \cos \theta) \left(\frac{2u \sin \theta}{g}\right) + \frac{1}{2} \left(\frac{g}{4}\right) \left(\frac{2u \sin \theta}{g}\right)^2 \]

### Step 6: Simplifying the New Range
1. The first term becomes:
\[ R' = \frac{2u^2 \sin \theta \cos \theta}{g} \]
This is equal to the original range \( R \).

2. The second term simplifies to:
\[ \frac{1}{2} \cdot \frac{g}{4} \cdot \frac{4u^2 \sin^2 \theta}{g^2} = \frac{u^2 \sin^2 \theta}{g} \]
This term is equal to \( H \) since:
\[ H = \frac{(u \sin \theta)^2}{2g} \]

### Step 7: Final Expression for New Range
Thus, the new range \( R' \) can be expressed as:
\[ R' = R + H \]

### Summary of Results
- The new horizontal range is \( R' = R + H \).
- The maximum height remains \( H \).

### Final Answer
- The new horizontal range is \( R + H \).
- The maximum height remains \( H \).