A particle is thrown from a stationary platform with velocity v at an angle of `60^(@)` with the horizontal. The range obtained is R. If the platform moves horizontally in the direction of target with velocity v, the range will increase to :

Options

A. `(3R)/2`
B. `(5R)/2`
C. `2 R`
D. `3 R`

Answer

To solve the problem, we need to analyze the motion of the particle thrown from a platform that is moving horizontally. Let's break down the solution step by step.

### Step 1: Determine the Range of the Particle from a Stationary Platform
When the particle is thrown from a stationary platform with an initial velocity \( v \) at an angle of \( 60^\circ \) with the horizontal, we can calculate the range \( R \) using the formula:

\[ R = \frac{v^2 \sin(2\theta)}{g} \]

Substituting \( \theta = 60^\circ \):

\[ R = \frac{v^2 \sin(120^\circ)}{g} \]

Since \( \sin(120^\circ) = \frac{\sqrt{3}}{2} \):

\[ R = \frac{v^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3} v^2}{2g} \]

### Step 2: Determine the Time of Flight
The time of flight \( T \) for the projectile can be calculated using the vertical component of the velocity:

\[ T = \frac{2v \sin(\theta)}{g} \]

Substituting \( \theta = 60^\circ \):

\[ T = \frac{2v \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3} v}{g} \]

### Step 3: Determine the Range when the Platform is Moving
When the platform moves horizontally with velocity \( v \), the horizontal velocity of the projectile becomes:

\[ v_{\text{horizontal}} = v \cos(60^\circ) + v \]

Since \( \cos(60^\circ) = \frac{1}{2} \):

\[ v_{\text{horizontal}} = v \cdot \frac{1}{2} + v = \frac{v}{2} + v = \frac{3v}{2} \]

Now, the new range \( R' \) when the platform moves is given by:

\[ R' = v_{\text{horizontal}} \cdot T \]

Substituting the values:

\[ R' = \left(\frac{3v}{2}\right) \cdot \left(\frac{\sqrt{3} v}{g}\right) = \frac{3\sqrt{3} v^2}{2g} \]

### Step 4: Calculate the Ratio of the New Range to the Original Range
To find how much the range increases, we take the ratio of the new range \( R' \) to the original range \( R \):

\[ \frac{R'}{R} = \frac{\frac{3\sqrt{3} v^2}{2g}}{\frac{\sqrt{3} v^2}{2g}} = \frac{3\sqrt{3}}{\sqrt{3}} = 3 \]

### Conclusion
Thus, the new range \( R' \) is three times the original range \( R \):

\[ R' = 3R \]

### Final Answer
The range will increase to \( 3R \).

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