Two particles A and B are projected with same speed so that the ratio of their maximum height reached is 3 : 1. If the speed of A is doubled without altering other parameters, the ratio of the horizontal ranges attained by A and B is

Options

A. `1:1`
B. `2:1`
C. `4:1`
D. `3:2`

Explanation

To solve the problem, we need to find the ratio of the horizontal ranges attained by particles A and B after doubling the speed of particle A. Let's break down the solution step by step.

### Step 1: Understand the Given Information
- Particles A and B are projected with the same initial speed \( U \).
- The ratio of their maximum heights is given as \( \frac{H_A}{H_B} = 3:1 \).

### Step 2: Relate Maximum Height to Projection Angle
The maximum height \( H \) of a projectile is given by the formula:
\[ H = \frac{U^2 \sin^2 \theta}{2g} \]
where \( \theta \) is the angle of projection and \( g \) is the acceleration due to gravity.

For particles A and B:
\[ H_A = \frac{U^2 \sin^2 \theta_1}{2g}, \quad H_B = \frac{U^2 \sin^2 \theta_2}{2g} \]
Given the ratio \( \frac{H_A}{H_B} = \frac{3}{1} \), we can write:
\[ \frac{\sin^2 \theta_1}{\sin^2 \theta_2} = 3 \]

### Step 3: Find the Angles of Projection
Let’s assume:
- \( \theta_1 = 60^\circ \) (for particle A)
- \( \theta_2 = 30^\circ \) (for particle B)

We can verify:
\[ \sin^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}, \quad \sin^2 30^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
Thus,
\[ \frac{\sin^2 60^\circ}{\sin^2 30^\circ} = \frac{\frac{3}{4}}{\frac{1}{4}} = 3 \]
This confirms our assumption about the angles of projection.

### Step 4: Calculate the Ranges of A and B
The range \( R \) of a projectile is given by:
\[ R = \frac{U^2 \sin 2\theta}{g} \]
For particle A:
\[ R_A = \frac{U^2 \sin 2\theta_1}{g} = \frac{U^2 \sin 120^\circ}{g} = \frac{U^2 \cdot \frac{\sqrt{3}}{2}}{g} \]
For particle B:
\[ R_B = \frac{U^2 \sin 2\theta_2}{g} = \frac{U^2 \sin 60^\circ}{g} = \frac{U^2 \cdot \frac{\sqrt{3}}{2}}{g} \]

### Step 5: Doubling the Speed of A
Now, if the speed of particle A is doubled, the new speed becomes \( 2U \). The new range for particle A is:
\[ R_A' = \frac{(2U)^2 \sin 120^\circ}{g} = \frac{4U^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{2\sqrt{3}U^2}{g} \]

### Step 6: Calculate the Ratio of Ranges
Now we find the ratio of the ranges of A and B:
\[ \frac{R_A'}{R_B} = \frac{\frac{2\sqrt{3}U^2}{g}}{\frac{\sqrt{3}U^2}{g}} = \frac{2\sqrt{3}}{\sqrt{3}} = 2 \]

### Final Answer
Thus, the ratio of the horizontal ranges attained by A and B after doubling the speed of A is:
\[ \frac{R_A'}{R_B} = 4:1 \]